NCERT Solutions for Class 12 Maths Chapter 13 Probability

Edited By Ramraj Saini | Updated on Mar 25, 2025 01:00 AM IST | #CBSE Class 12th

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CBSE Class 12th Exam Date : 03 Apr' 2025 - 03 Apr' 2025

Everyday before stepping out of our houses, we check for weather forecasts. Ever wondered how metrologists predict the weather? Ever thought about how these clothing brands determine what’s going to trend or on what models, AI and machine learning work? The solution is in the Probability chapter! A basic and significant topic in NCERT Class 12 Maths is probability. It exposes students to the concepts of probability distributions, random variables, Bayes' theorem, and conditional probability. Because the questions follow the same patterns as provided in the NCERT Solutions, students getting ready for competitive exams are recommended to practice their worksheets and exercises consistently, therefore enabling them to establish a strong basis for both exams and practical uses.

Students can additionally refer to NCERT Exemplar Solutions For Class 12 Maths Chapter 13 Probability for better practice and understanding of the topic.

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NCERT Probability Class 12 Questions And Answers PDF Free Download
NCERT Class 12 Maths Chapter 13 Question Answer Probability - Important Formulae
NCERT Probability Class 12 Questions And Answers (Exercise)
Importance of solving NCERT questions for class 12 Math Chapter 13
NCERT Exemplar Class 12 Solutions
NCERT solutions for class 12 subject wise
NCERT Solutions class wise
NCERT Books and NCERT Syllabus

NCERT Probability Class 12 Questions And Answers PDF Free Download

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NCERT Class 12 Maths Chapter 13 Question Answer Probability - Important Formulae

Conditional Probability: Conditional probability is the likelihood of an event occurring based on the occurrence of a preceding event. For two events A and B with the same sample space, the conditional probability of event A given that B has occurred (P(A|B)) is defined as:

$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ (when P(B) ≠ 0)

Other conditional probability relationships:

$\begin{aligned} P (S ∣ F) = P (F ∣ F) = 1 \\ P ((A \cup B) ∣ F) = P (A ∣ F) + P (B ∣ F) - P ((A \cap B) ∣ F) \\ P (E^{'} ∣ F) = 1 - P (E ∣ F) \end{aligned}$

Multiplication Rule: The multiplication rule relates the probability of two events E and F in a sample space S:

$P (E \cap F) = P (E) \cdot P (F ∣ E) = P (F) \cdot P (E ∣ F)$ (when P(E) ≠ 0 and P(F) ≠ 0)

Independent Events: Two experiments are considered independent if the probability of the events E and F occurring simultaneously is the product of their individual probabilities:

$P (E \cap F) = P (E) \cdot P (F)$

Bayes’ Theorem: Bayes’ theorem deals with events E1, E2, …, En that form a partition of the sample space S. It allows the calculation of the probability of event Ei given event A:

$P (E_{i} ∣ A) = \frac{P (E_{i}) \cdot P (A ∣ E_{i})}{\sum_{j = 1}^{n} P (E_{j}) \cdot P (A ∣ E_{j})}$ , for i = 1, 2, …, n

Theorem of Total Probability: Given a partition E1, E2, …, En of the sample space and an event A, the theorem of total probability states:

$P (A) = P (E_{1}) \cdot P (A ∣ E_{1}) + P (E_{2}) \cdot P (A ∣ E_{2}) + \dots + P (E_{n}) \cdot P (A ∣ E_{n})$

Free download NCERT Class 12 Maths Chapter 13 Question Answer Probability for CBSE Exam.

NCERT Probability Class 12 Questions And Answers (Exercise)

Class 12 Maths chapter 13 solutions - Exercise: 13.1
Page number: 413-415
Total questions: 17

Question:1 Given that $E$ and $F$ are events such that $P (E) = 0.6, P (F) = 0.3$ and $p (E \cap F) = 0.2,$ find $P (E ∣ F)$ and $P (F ∣ E)$

Answer:

It is given that $P (E) = 0.6, P (F) = 0.3$ and $p (E \cap F) = 0.2,$

$P (E | F) = \frac{p (E \cap F)}{P (F)} = \frac{0.2}{0.3} = \frac{2}{3}$

$P (F | E) = \frac{p (E \cap F)}{P (E)} = \frac{0.2}{0.6} = \frac{1}{3}$

Question:2 Compute $P (A ∣ B),$ if $P (B) = 0.5$ and $P (A \cap B) = 0.32$

Answer:

It is given that $P (B) = 0.5$ and $P (A \cap B) = 0.32$

$P (A | B) = \frac{p (A \cap B)}{P (B)} = \frac{0.32}{0.5} = 0.64$

Question:3 If $P (A) = 0.8, P (B) = 0.5$ and $P (B ∣ A) = 0.4,$ find

(i) $P (A \cap B)$

(ii) $P (A ∣ B)$

(iii) $P (A \cup B)$

Answer:

(i) It is given that P(A)=0.8, P(B)=0.5 and P(B|A)=0.4

$P (B | A) = \frac{p (A \cap B)}{P (A)}$

$\Rightarrow$ $0.4 = \frac{p (A \cap B)}{0.8}$

$\Rightarrow$ $p (A \cap B) = 0.4 \times 0.8$

$\Rightarrow$ $p (A \cap B) = 0.32$

(ii) It is given that $P (A) = 0.8, P (B) = 0.5$ and $P (B ∣ A) = 0.4,$

$P (A \cap B) = 0.32$

$P (A | B) = \frac{p (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{0.32}{0.5}$

$\Rightarrow$ $P (A | B) = \frac{32}{50} = 0.64$

(iii) It is given that $P (A) = 0.8, P (B) = 0.5$

$P (A \cap B) = 0.32$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $P (A \cup B) = 0.8 + 0.5 - 0.32$

$\Rightarrow$ $P (A \cup B) = 1.3 - 0.32$

$\Rightarrow$ $P (A \cup B) = 0.98$

Question:4 Evaluate $P (A \cup B),$ if $2 P (A) = P (B) = \frac{5}{13}$ and $P (A ∣ B) = \frac{2}{5}$

Answer:

Given in the question $2 P (A) = P (B) = \frac{5}{13}$ and $P (A ∣ B) = \frac{2}{5}$

We know that:

$P (A | B) = \frac{p (A \cap B)}{P (B)}$

$\Rightarrow$ $\frac{2}{5} = \frac{p (A \cap B)}{\frac{5}{13}}$

$\Rightarrow$ $\frac{2 \times 5}{5 \times 13} = p (A \cap B)$

$\Rightarrow$ $p (A \cap B) = \frac{2}{13}$

Use, $p (A \cup B) = p (A) + p (B) - p (A \cap B)$

$\Rightarrow$ $p (A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$

$\Rightarrow$ $p (A \cup B) = \frac{11}{26}$

Question:5 If $P (A) = \frac{6}{11}, P (B) = \frac{5}{11}$ and $P (A \cup B) = \frac{7}{11} .$ , find

(i) $P (A \cap B)$

(ii) $P (A ∣ B)$

(iii) $P (B ∣ A)$

Answer:

(i) Given in the question

$P (A) = \frac{6}{11}, P (B) = \frac{5}{11}$ and $P (A \cup B) = \frac{7}{11} .$

By using formula:

$p (A \cup B) = p (A) + p (B) - p (A \cap B)$

$\Rightarrow$ $\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - p (A \cap B)$

$\Rightarrow$ $p (A \cap B) = \frac{11}{11} - \frac{7}{11}$

$\Rightarrow$ $p (A \cap B) = \frac{4}{11}$

(ii) It is given that - $P (A) = \frac{6}{11}, P (B) = \frac{5}{11}$

$p (A \cap B) = \frac{4}{11}$

We know that:

$P (A | B) = \frac{p (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{4}{11}}{\frac{5}{11}}$

$\Rightarrow$ $P (A | B) = \frac{4}{5}$

(iii) Given in the question-

$P (A) = \frac{6}{11}, P (B) = \frac{5}{11}$ and $p (A \cap B) = \frac{4}{11}$

Use formula

$P (B | A) = \frac{p (A \cap B)}{P (A)}$

$\Rightarrow$ $P (B | A) = \frac{\frac{4}{11}}{\frac{6}{11}}$

$\Rightarrow$ $P (B | A) = \frac{4}{6} = \frac{2}{3}$

Question:6 A coin is tossed three times, where

(i) E: head on third toss,F: heads on first two tosses

(ii) E: at least two heads,F: at most two heads

(iii) E: at most two tails,F: at least one tail

Answer:

(i) The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes $= 2^{3} = 8$

According to question

E: head on third toss, F: heads on first two tosses

$E = {H H H, T T H, H T H, T H H}$

$F = {H H H, H H T}$

$E \cap F = H H H$

$P (F) = \frac{2}{8} = \frac{1}{4}$

$P (E \cap F) = \frac{1}{8}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{1}{8}}{\frac{1}{4}}$

$\Rightarrow$ $P (E | F) = \frac{4}{8} = \frac{1}{2}$

(ii) The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes $= 2^{3} = 8$

According to question

E : at least two heads , F : at most two heads

$E = {H H H, H T H, T H H, H H T} = 4$

$F = {H T H, H H T, T H H, T T T, H T T, T T H, T H T} = 7$

$E \cap F = {H T H, T H H, H H T} = 3$

$P (F) = \frac{7}{8}$

$P (E \cap F) = \frac{3}{8}$

$\Rightarrow$ $P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{3}{8}}{\frac{7}{8}}$

$P (E | F) = \frac{3}{7}$

(iii) The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes $= 2^{3} = 8$

According to question

E: at most two tails, F: at least one tail

$E = {H H H, T T H, H T H, T H H, T H T, H T T, H H T} = 7$

$F = {T T T, T T H, H T H, T H H, T H T, H T T, H H T} = 7$

$E \cap F = {T T H, H T H, T H H, T H T, H T T, H H T} = 6$

$P (F) = \frac{7}{8}$

$P (E \cap F) = \frac{6}{8} = \frac{3}{4}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{3}{4}}{\frac{7}{8}}$

$\Rightarrow$ $P (E | F) = \frac{6}{7}$

Question:7 Two coins are tossed once, where

(i) E : tail appears on one coin, F : one coin shows head

(ii) E : no tail appears,F : no head appears

Answer:

(i) E : tail appears on one coin, F : one coin shows head

Total outcomes =4

$E = {H T, T H} = 2$

$F = {H T, T H} = 2$

$E \cap F = {H T, T H} = 2$

$P (F) = \frac{2}{4} = \frac{1}{2}$

$P (E \cap F) = \frac{2}{4} = \frac{1}{2}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{1}{2}}{\frac{1}{2}}$

$\Rightarrow$ $P (E | F) = 1$

(ii) E : no tail appears, F : no head appears

Total outcomes =4

$E = H H F = T T$

$E \cap F = ϕ$

$n (E \cap F) = 0$

$P (F) = 1$

$P (E \cap F) = \frac{0}{4} = 0$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$P (E | F) = \frac{0}{1} = 0$

Question:8 A die is thrown three times,

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Answer:

E: 4 appears on the third toss, F: 6 and 5 appear respectively on the first two tosses

Total outcomes $= 6^{3} = 216$

$n (E) = 36$ as 4 is fixed at third place.

$F = {651, 652, 653, 654, 655, 656}$

$n (F) = 6$

$E \cap F = {654}$

$n (E \cap F) = 1$

$P (E \cap F) = \frac{1}{216}$

$P (F) = \frac{6}{216} = \frac{1}{36}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{1}{216}}{\frac{1}{36}}$

$\Rightarrow$ $P (E | F) = \frac{1}{6}$

Question:9 Mother, father and son line up at random for a family picture

E : son on one end, F : father in middle

Answer:

E: son on one end, F: father in middle

Total outcomes $= 3! = 3 \times 2 = 6$

Let S be son, M be mother and F be father.

Then we have,

$E = {S M F, S F M, F M S, M F S}$

$n (E) = 4$

$F = {S F M, M F S}$

$n (F) = 2$

$E \cap F = {S F M, M F S}$

$n (E \cap F) = 2$

$P (F) = \frac{2}{6} = \frac{1}{3}$

$P (E \cap F) = \frac{2}{6} = \frac{1}{3}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{1}{3}}{\frac{1}{3}}$

$\Rightarrow$ $P (E | F) = 1$

Question:10 A black and a red dice are rolled.

(i) Find the conditional probability of obtaining a sum greater than $9$ , given that the black die resulted in a $5.$

(ii) Find the conditional probability of obtaining the sum $8$ , given that the red die resulted in a number less than $4$ .

Answer:

(i) A black and a red dice are rolled.

Total outcomes $= 6^{2} = 36$

Let the A be an event obtaining a sum greater than $9$ and B be an event that the black die resulted in a $5.$

$A = {46, 55, 56, 64, 65, 66}$

$n (A) = 6$

$B = {51, 52, 53, 54, 55, 56}$

$n (B) = 6$

$A \cap B = {55, 56}$

$n (A \cap B) = 2$

$P (A \cap B) = \frac{2}{36}$

$P (B) = \frac{6}{36}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$P (A | B) = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3}$

(ii) A black and a red dice are rolled.

Total outcomes $= 6^{2} = 36$

Let the A be event obtaining a sum 8 and B be a event thatthat the red die resulted in a number less than $4$ .

$A = {26, 35, 53, 44, 62,}$

$n (A) = 5$

Red dice is rolled after black dice.

$B = {11, 12, 13, 21, 22, 23, 31, 32, 33, 41, 42, 43, 51, 52, 53, 61, 62, 63}$

$n (B) = 18$

$A \cap B = {53, 62}$

$n (A \cap B) = 2$

$P (A \cap B) = \frac{2}{36}$

$P (B) = \frac{18}{36}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$P (A | B) = \frac{\frac{2}{36}}{\frac{18}{36}} = \frac{2}{18} = \frac{1}{9}$

Question:11 A fair die is rolled. Consider events $E = {1, 3, 5}, F {2, 3}$ and $G = {2, 3, 4, 5}$ Find

(i) $P (E ∣ F)$ and $P (F ∣ E)$

(ii) $P (E ∣ G)$ and $P (G ∣ E)$

(iii) $P ((E \cup F) ∣ G)$ and $P ((E \cap F) ∣ G)$

Answer:

(i) A fair die is rolled.

Total oucomes $= {1, 2, 3, 4, 5, 6} = 6$

$E = {1, 3, 5}, F {2, 3}$

$E \cap F = {3}$

$n (E \cap F) = 1$

$n (F) = 2$

$n (E) = 3$

$P (E) = \frac{3}{6}$ $P (F) = \frac{2}{6}$ and $P (E \cap F) = \frac{1}{6}$

$P (E | F) = \frac{P (E \cap F)}{P (F)}$

$\Rightarrow$ $P (E | F) = \frac{\frac{1}{6}}{\frac{2}{6}}$

$\Rightarrow$ $P (E | F) = \frac{1}{2}$

$P (F | E) = \frac{P (F \cap E)}{P (E)}$

$\Rightarrow$ $P (F | E) = \frac{\frac{1}{6}}{\frac{3}{6}}$

$\Rightarrow$ $P (F | E) = \frac{1}{3}$

(ii) A fair die is rolled.

Total oucomes $= {1, 2, 3, 4, 5, 6} = 6$

$E = {1, 3, 5}$ , $G = {2, 3, 4, 5}$

$E \cap G = {3, 5}$

$n (E \cap G) = 2$

$n (G) = 4$

$n (E) = 3$

$P (E) = \frac{3}{6}$ $P (G) = \frac{4}{6}$ $P (E \cap F) = \frac{2}{6}$

$P (E | G) = \frac{P (E \cap G)}{P (G)}$

$\Rightarrow$ $P (E | G) = \frac{\frac{2}{6}}{\frac{4}{6}}$

$\Rightarrow$ $P (E | G) = \frac{2}{4} = \frac{1}{2}$

$P (G | E) = \frac{P (G \cap E)}{P (E)}$

$\Rightarrow$ $P (G | E) = \frac{\frac{2}{6}}{\frac{3}{6}}$

$\Rightarrow$ $P (G | E) = \frac{2}{3}$

(iii) A fair die is rolled.

Total oucomes $= {1, 2, 3, 4, 5, 6} = 6$

$E = {1, 3, 5}, F {2, 3}$ and $G = {2, 3, 4, 5}$

$E \cap G = {3, 5}$ , $F \cap G = {2, 3}$

$(E \cap G) \cap G = {3}$

$P [(E \cap G) \cap G] = \frac{1}{6}$ $P (E \cap G) = \frac{2}{6}$ $P (F \cap G) = \frac{2}{6}$

$P ((E \cup F) | G) = P (E | G) + P (F | G) - P [(E \cap F) | G]$

$\Rightarrow$ $= \frac{P (E \cap G)}{P (G)} + \frac{P (F \cap G)}{P (G)} - \frac{P ((E \cap F) \cap G)}{P (G)}$

$\Rightarrow$ $= \frac{\frac{2}{6}}{\frac{4}{6}} + \frac{\frac{2}{6}}{\frac{4}{6}} - \frac{\frac{1}{6}}{\frac{4}{6}}$

$\Rightarrow$ $= \frac{2}{4} + \frac{2}{4} - \frac{1}{4}$

$\Rightarrow$ $= \frac{3}{4}$

$P ((E \cap F) | G) = \frac{P ((E \cap F) \cap G)}{P (G)}$

$\Rightarrow$ $P ((E \cap F) | G) = \frac{\frac{1}{6}}{\frac{4}{6}}$

$\Rightarrow$ $P ((E \cap F) | G) = \frac{1}{4}$

Question:12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(i) the youngest is a girl,

(ii) at least one is a girl?

Answer:

(i) Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by $G 1 a n d G 2$ respectively also first and second boy are denoted by $B 1 a n d B 2$

If a family has two children, then total outcomes $= 2^{2} = 4$ $= {(B 1 B 2), (G 1 G 2), (G 1 B 2), (G 2 B 1)}$

Let A= both are girls $= {(G 1 G 2)}$

and B= the youngest is a girl = $= {(G 1 G 2), (B 1 G 2)}$

$A \cap B = {(G 1 G 2)}$

$P (A \cap B) = \frac{1}{4}$ $P (B) = \frac{2}{4}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{1}{4}}{\frac{2}{4}}$

$\Rightarrow$ $P (A | B) = \frac{1}{2}$

Therefore, the required probability is 1/2

(ii) Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by $G 1 a n d G 2$ respectively also first and second boy are denoted by $B 1 a n d B 2$

If a family has two children, then total outcomes $= 2^{2} = 4$ $= {(B 1 B 2), (G 1 G 2), (G 1 B 2), (G 2 B 1)}$

Let A= both are girls $= {(G 1 G 2)}$

and C= at least one is a girl = $= {(G 1 G 2), (B 1 G 2), (G 1 B 2)}$

$A \cap B = {(G 1 G 2)}$

$P (A \cap B) = \frac{1}{4}$ $P (C) = \frac{3}{4}$

$P (A | C) = \frac{P (A \cap C)}{P (C)}$

$\Rightarrow$ $P (A | C) = \frac{\frac{1}{4}}{\frac{3}{4}}$

$\Rightarrow$ $P (A | C) = \frac{1}{3}$

Question:13 An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer:

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.

Total number of questions $= 300 + 200 + 500 + 400 = 1400$

Let A = question be easy.

$n (A) = 300 + 500 = 800$

$P (A) = \frac{800}{1400} = \frac{8}{14}$

Let B = multiple choice question

$n (B) = 500 + 400 = 900$

$P (B) = \frac{900}{1400} = \frac{9}{14}$

$A \cap B =$ easy multiple questions

$n (A \cap B) = 500$

$P (A \cap B) = \frac{500}{1400} = \frac{5}{14}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{5}{14}}{\frac{9}{14}}$

$\Rightarrow$ $P (A | B) = \frac{5}{9}$

Therefore, the required probability is 5/9

Question:14 Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer:

Two dice are thrown.

Total outcomes $= 6^{2} = 36$

Let A be the event ‘the sum of numbers on the dice is 4.

$A = {(13), (22), (31)}$

Let B be the event that two numbers appearing on throwing two dice are different.

$B = {(12), (13), (14), (15), (16), (21) (23), (24), (25), (26,) (31), (32), (34), (35), (36), (41), (42), (43), (45), (46), (51), (52), (53), (54), (56), (61), (62), (63), (64), (65)}$ $n (B) = 30$

$P (B) = \frac{30}{36}$

$A \cap B = {(13), (31)}$

$n (A \cap B) = 2$

$P (A \cap B) = \frac{2}{36}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{2}{36}}{\frac{30}{36}}$

$\Rightarrow$ $P (A | B) = \frac{2}{30} = \frac{1}{15}$

Therefore, the required probability is 1/15

Question:15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin.

Total outcomes $= {(1 H), (1 T), (2 H), (2 T), (31), (32), (33), (34), (35), (36), (4 H), (4 T), (5 H), (5 T), (61), (62), (63), (64), (65), (66)}$

Total number of outcomes =20

Let A be a event when coin shows a tail.

$A = {((1 T), (2 T), (4 T), (5 T)}$

Let B be a event that ‘at least one die shows a 3’.

$B = {(31), (32), (33), (34), (35), (36), (63)}$

$n (B) = 7$

$P (B) = \frac{7}{20}$

$A \cap B = ϕ$

$n (A \cap B) = 0$

$P (A \cap B) = \frac{0}{20} = 0$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{0}{\frac{7}{20}}$

$\Rightarrow$ $P (A | B) = 0$

Question:16 In the following Exercise 16 choose the correct answer:

If $P (A) = \frac{1}{2}, P (B) = 0,$ then $P (A ∣ B)$ is

(A) $0$

(B) $\frac{1}{2}$

(D) $1$

Answer:

It is given that

$P (A) = \frac{1}{2}, P (B) = 0,$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{P (A \cap B)}{0}$

Hence, P (A|B) is not defined.

Thus, the correct option is C.

Question:17 In the following Exercise 17 choose the correct answer:

If $A$ and $B$ are events such that $P (A ∣ B) = P (B ∣ A),$ then

(A) $A \subset B$ but $A \neq B$

(B) $A = B$

(D) $P (A) = P (B)$

Answer:

It is given that $P (A ∣ B) = P (B ∣ A),$

$\Rightarrow$ $\frac{P (A \cap B)}{P (B)}$ $= \frac{P (A \cap B)}{P (A)}$

$\Rightarrow$ $P (A) = P (B)$

Hence, option D is correct.

Class 12 Maths chapter 13 solutions - Exercise: 13.2
Page number: 421-423
Total questions: 18

Question:1 If $P (A) = \frac{3}{5}$ and $P (B) = \frac{1}{5},$ find $P (A \cap B)$ if $A$ and $B$ are independent events.

Answer:

$P (A) = \frac{3}{5}$ and $P (B) = \frac{1}{5},$

Given : $A$ and $B$ are independent events.

So we have, $P (A \cap B) = P (A) . P (B)$

$\Rightarrow P (A \cap B) = \frac{3}{5} \times \frac{1}{5}$

$\Rightarrow P (A \cap B) = \frac{3}{25}$

Question:2 Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer:

Two cards are drawn at random and without replacement from a pack of 52 playing cards.

There are 26 black cards in a pack of 52.

Let $P (A)$ be the probability that first cards is black.

Then, we have

$P (A) = \frac{26}{52} = \frac{1}{2}$

Let $P (B)$ be the probability that second cards is black.

Then, we have

$P (B) = \frac{25}{51}$

The probability that both the cards are black $= P (A) . P (B)$

$\Rightarrow$ $= \frac{1}{2} \times \frac{25}{51}$

$\Rightarrow$ $= \frac{25}{102}$

Question:3 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.

Answer:

Total oranges = 15

Good oranges = 12

Bad oranges = 3

Let $P (A)$ be the probability that first orange is good.

The, we have

$P (A) = \frac{12}{15} = \frac{4}{5}$

Let $P (B)$ be the probability that second orange is good.

$P (B) = \frac{11}{14}$

Let $P (C)$ be the probability that third orange is good.

$P (C) = \frac{10}{13}$

The probability that a box will be approved for sale $= P (A) . P (B) . P (C)$

$\Rightarrow$ $\frac{4}{5} . \frac{11}{14} . \frac{10}{13}$

$\Rightarrow$ $\frac{44}{91}$

Question:4 A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Answer:

A fair coin and an unbiased die are tossed,then total outputs are:

$= {(H 1), (H 2), (H 3), (H 4), (H 5), (H 6), (T 1), (T 2), (T 3), (T 4), (T 5), (T 6)}$

$= 12$

A is the event ‘head appears on the coin’ .

Total outcomes of A are : $= {(H 1), (H 2), (H 3), (H 4), (H 5), (H 6)}$

$P (A) = \frac{6}{12} = \frac{1}{2}$

B is the event ‘3 on the die’.

Total outcomes of B are : $= {(T 3), (H 3)}$

$P (B) = \frac{2}{12} = \frac{1}{6}$

$∴ A \cap B = (H 3)$

$P (A \cap B) = \frac{1}{12}$

Also, $P (A \cap B) = P (A) . P (B)$

$P (A \cap B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Hence, A and B are independent events.

Question:5 A die marked $1, 2, 3$ in red and $4, 5, 6$ in green is tossed. Let $A$ be the event, ‘the number is even,’ and $B$ be the event, ‘the number is red’. Are $A$ and $B$ independent?

Answer:

Total outcomes $= {1, 2, 3, 4, 5, 6} = 6$ .

$A$ is the event, ‘the number is even,’

Outcomes of A $= {2, 4, 6}$

$n (A) = 3.$

$P (A) = \frac{3}{6} = \frac{1}{2}$

$B$ is the event, ‘the number is red’.

Outcomes of B $= {1, 2, 3}$

$n (B) = 3.$

$P (B) = \frac{3}{6} = \frac{1}{2}$

$∴ (A \cap B) = {2}$

$n (A \cap B) = 1$

$P (A \cap B) = \frac{1}{6}$

Also,

$P (A \cap B) = P (A) . P (B)$

$P (A \cap B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \neq \frac{1}{6}$

Thus, both the events A and B are not independent.

Question:6 Let $E$ and $F$ be events with $P (E) = \frac{3}{5}, P (F) = \frac{3}{10}$ and $P (E \cap F) = \frac{1}{5} .$ Are E and F independent?

Answer:

Given :

$P (E) = \frac{3}{5}, P (F) = \frac{3}{10}$ and $P (E \cap F) = \frac{1}{5} .$

For events E and F to be independent , we need

$P (E \cap F) = P (E) . P (F)$

$P (E \cap F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50} \neq \frac{1}{5}$

Hence, E and F are not indepent events.

Question:7 Given that the events $A$ and $B$ are such that $P (A) = \frac{1}{2}, P (A \cup B) = \frac{3}{5}$ and $P (B) = p .$ Find $p$ if they are

(i) mutually exclusive

(ii) independent

Answer:

(i) Given,

$P (A) = \frac{1}{2}, P (A \cup B) = \frac{3}{5}$

Also, A and B are mutually exclusive means $A \cap B = ϕ$ .

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $\frac{3}{5} = \frac{1}{2} + P (B) - 0$

$\Rightarrow$ $P (B) = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$

(ii) Given,

$P (A) = \frac{1}{2}, P (A \cup B) = \frac{3}{5}$

Also, A and B are independent events means

$P (A \cap B) = P (A) . P (B)$ . Also $P (B) = p .$

$P (A \cap B) = P (A) . P (B) = \frac{p}{2}$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}$

$\Rightarrow$ $\frac{p}{2} = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$

$\Rightarrow$ $p = \frac{2}{10} = \frac{1}{5}$

Question:8 Let A and B be independent events with $P (A) = 0.3$ and $P (B) = 0.4$ Find

(i) $P (A \cap B)$

(ii) $P (A \cup B)$

(iii) $P (A ∣ B)$

(iv) $P (B ∣ A)$

Answer:

(i) $P (A) = 0.3$ and $P (B) = 0.4$

Given : A and B be independent events

So, we have

$P (A \cap B) = P (A) . P (B)$

$\Rightarrow$ $P (A \cap B) = 0.3 \times 0.4 = 0.12$

(ii) $P (A) = 0.3$ and $P (B) = 0.4$

Given : A and B be independent events

So, we have

$P (A \cap B) = P (A) . P (B)$

$P (A \cap B) = 0.3 \times 0.4 = 0.12$

We have, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $P (A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$

(iii) $P (A) = 0.3$ and $P (B) = 0.4$

Given : A and B be independent events

So, we have $P (A \cap B) = 0.12$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{0.12}{0.4} = 0.3$

(iv) $P (A) = 0.3$ and $P (B) = 0.4$

Given : A and B be independent events

So, we have $P (A \cap B) = 0.12$

$P (B | A) = \frac{P (A \cap B)}{P (A)}$

$\Rightarrow$ $P (B | A) = \frac{0.12}{0.3} = 0.4$

Question:9 If $A$ and $B$ are two events such that $P (A) = \frac{1}{4}, P (B) = \frac{1}{2}$ and $P (A \cap B) = \frac{1}{8},$ find $P (n o t A a n d n o t B) .$

Answer:

If $A$ and $B$ are two events such that $P (A) = \frac{1}{4}, P (B) = \frac{1}{2}$ and $P (A \cap B) = \frac{1}{8},$

$P (n o t A a n d n o t B) = P (A^{'} \cap B^{'})$

$P (n o t A a n d n o t B) = P (A \cup B)^{'}$ use, $(P (A^{'} \cap B^{'}) = P (A \cup B)^{'})$

$\Rightarrow$ $1 - (P (A) + P (B) - P (A \cap B))$

$\Rightarrow$ $1 - (\frac{1}{4} + \frac{1}{2} - \frac{1}{8})$

$\Rightarrow$ $1 - (\frac{6}{8} - \frac{1}{8})$

$\Rightarrow$ $1 - \frac{5}{8}$

$\Rightarrow$ $\frac{3}{8}$

Question:10 Events A and B are such that $P (A) = \frac{1}{2}, P (B) = \frac{7}{12}$ and $P (n o t A o r n o t B) = \frac{1}{4} .$ State whether $A$ and $B$ are independent ?

Answer:

If $A$ and $B$ are two events such that $P (A) = \frac{1}{2}, P (B) = \frac{7}{12}$ and $P (n o t A o r n o t B) = \frac{1}{4} .$

$P (A^{'} \cup B^{'}) = \frac{1}{4}$

$P (A \cap B)^{'} = \frac{1}{4}$ $(A^{'} \cup B^{'} = (A \cap B)^{'})$

$\Rightarrow 1 - P (A \cap B) = \frac{1}{4}$

$\Rightarrow P (A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$

$A l s o P (A \cap B) = P (A) . P (B)$

$P (A \cap B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$

As we can see $\frac{3}{4} \neq \frac{7}{24}$

Hence, A and B are not independent.

Question:11 Given two independent events $A$ and $B$ such that $P (A) = 0.3, P (B) = 0.6,$ Find

(i) $P (A a n d B)$

(ii) $P (A a n d n o t B)$

(iii) $P (A o r B)$

(iv) $P (n e i t h e r A n o r B)$

Answer:

(i) $P (A) = 0.3, P (B) = 0.6,$

Given two independent events $A$ and $B$ .

$P (A \cap B) = P (A) . P (B)$

$\Rightarrow$ $P (A \cap B) = 0.3 \times 0.6 = 0.18$

Also , we know $P (A a n d B) = P (A \cap B) = 0.18$

(ii) $P (A) = 0.3, P (B) = 0.6,$

Given two independent events $A$ and $B$ .

$P (A a n d n o t B)$ $= P (A) - P (A \cap B)$

$0.3 - 0.18 = 0.12$

(iii) $P (A) = 0.3, P (B) = 0.6,$

$P (A \cap B) = 0.18$

$P (A o r B) = P (A \cup B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $0.3 + 0.6 - 0.18$

$\Rightarrow$ $0.9 - 0.18$

$\Rightarrow$ $0.72$

(iv) $P (A) = 0.3, P (B) = 0.6,$

$P (A \cap B) = 0.18$

$P (A o r B) = P (A \cup B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $0.3 + 0.6 - 0.18$

$\Rightarrow$ $0.9 - 0.18$

$\Rightarrow$ $0.72$

$P (n e i t h e r A n o r B)$ $= P (A^{'} \cap B^{'})$

$= P ((A \cup B)^{'})$

$= 1 - P (A \cup B)$

$1 - 0.72$

$0.28$

Question:12 A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer:

A die is tossed thrice.

Outcomes $= {1, 2, 3, 4, 5, 6}$

Odd numbers $= {1, 3, 5}$

The probability of getting an odd number at first throw

$= \frac{3}{6} = \frac{1}{2}$

The probability of getting an even number

$= \frac{3}{6} = \frac{1}{2}$

Probability of getting even number three times

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

The probability of getting an odd number at least once = 1 - the probability of getting an odd number in none of throw

= 1 - probability of getting even number three times

$\Rightarrow$ $1 - \frac{1}{8}$

$\Rightarrow$ $\frac{7}{8}$

Question:13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red.

Answer:

(i) Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a red ball in first draw

$= \frac{8}{18} = \frac{4}{9}$

The ball is repleced after drawing first ball.

The probability of getting a red ball in second draw

$= \frac{8}{18} = \frac{4}{9}$

the probability that both balls are red

$= \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$

(ii) Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a black ball in the first draw

$= \frac{10}{18} = \frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$= \frac{8}{18} = \frac{4}{9}$

the probability that the first ball is black and the second is red

$= \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$

(iii) Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw

$= \frac{10}{18} = \frac{5}{9}$

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

$= \frac{8}{18} = \frac{4}{9}$

the probability that the first ball is black and the second is red

$= \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . .1$

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw

$= \frac{8}{18} = \frac{4}{9}$

The probability of getting a black ball in the second draw

$= \frac{10}{18} = \frac{5}{9}$

the probability that the first ball is red and the second is black

$= \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . .2$

Thus,

The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probability that the first ball is red and the second is black $= \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$

Question:14 Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved

(ii) exactly one of them solves the problem

Answer:

(i) $P (A) = \frac{1}{2}$ and $P (B) = \frac{1}{3}$

Since, problem is solved independently by A and B,

$∴$ $P (A \cap B) = P (A) . P (B)$

$\Rightarrow$ $P (A \cap B) = \frac{1}{2} \times \frac{1}{3}$

$\Rightarrow$ $P (A \cap B) = \frac{1}{6}$

probability that the problem is solved $= P (A \cup B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow$ $P (A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6}$

$\Rightarrow$ $P (A \cup B) = \frac{5}{6} - \frac{1}{6}$

$\Rightarrow$ $P (A \cup B) = \frac{4}{6} = \frac{2}{3}$

(ii) $P (A) = \frac{1}{2}$ and $P (B) = \frac{1}{3}$

$P (A^{'}) = 1 - P (A)$ , $P (B^{'}) = 1 - P (B)$

$P (A^{'}) = 1 - \frac{1}{2} = \frac{1}{2}$ , $P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$

probability that exactly one of them solves the problem $= P (A \cap B^{'}) + P (A^{'} \cap B)$

probability that exactly one of them solves the problem $= P (A) . P (B^{'}) + P (A^{'}) P (B)$

$\Rightarrow$ $\frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3}$

$\Rightarrow$ $\frac{2}{6} + \frac{1}{6}$

$\Rightarrow$ $\frac{3}{6} = \frac{1}{2}$

Question:15 One card is drawn at random from a well-shuffled deck of $52$ cards. In which of the following cases are the events $E$ and $F$ independent?

(i) E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

(ii) E : ‘the card drawn is

F : ‘the card drawn is a king’

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

Answer:

(i) One card is drawn at random from a well shuffled deck of $52$ cards

Total ace = 4

total spades =13

E : ‘the card drawn is a spade

F : ‘the card drawn is an ace’

$P (E) = \frac{13}{52} = \frac{1}{4}$

$P (F) = \frac{4}{52} = \frac{1}{13}$

$E \cap F :$ a card which is spade and ace = 1

$P (E \cap F) = \frac{1}{52}$

$P (E) . P (F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}$

$\Rightarrow P (E \cap F) = P (E) . P (F) = \frac{1}{52}$

Hence, E and F are independent events.

(ii) One card is drawn at random from a well shuffled deck of $52$ cards

Total black card = 26

total king =4

E : ‘the card drawn is black’

F : ‘the card drawn is a king’

$P (E) = \frac{26}{52} = \frac{1}{2}$

$P (F) = \frac{4}{52} = \frac{1}{13}$

$E \cap F :$ a card which is black and king = 2

$P (E \cap F) = \frac{2}{52} = \frac{1}{26}$

$P (E) . P (F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}$

$\Rightarrow P (E \cap F) = P (E) . P (F) = \frac{1}{26}$

Hence, E and F are independent events .

(iii) One card is drawn at random from a well shuffled deck of $52$ cards

Total king or queen = 8

total queen or jack = 8

E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

$P (E) = \frac{8}{52} = \frac{2}{13}$

$P (F) = \frac{8}{52} = \frac{2}{13}$

$E \cap F :$ a card which is queen = 4

$P (E \cap F) = \frac{4}{52} = \frac{1}{13}$

$P (E) . P (F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169}$

$\Rightarrow P (E \cap F) \neq P (E) . P (F)$

Hence, E and F are not independent events

Question:16 In a hostel, $60^{o} /_{o}$ of the students read Hindi newspaper, $40^{o} /_{o}$ read English newspaper and $20^{o} /_{o}$ read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Answer:

(a) H : $60^{o} /_{o}$ of the students read Hindi newspaper,

E : $40^{o} /_{o}$ read English newspaper and

$H \cap E :$ $20^{o} /_{o}$ read both Hindi and English newspapers.

$P (H) = \frac{60}{100} = \frac{6}{10} = \frac{3}{5}$

$P (E) = \frac{40}{100} = \frac{4}{10} = \frac{2}{5}$

$P (H \cap E) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$

the probability that she reads neither Hindi nor English newspapers $= 1 - P (H \cup E)$

$= 1 - (P (H) + P (E) - P (H \cap E))$

$\Rightarrow$ $1 - (\frac{3}{5} + \frac{2}{5} - \frac{1}{5})$

$\Rightarrow$ $1 - \frac{4}{5}$

$\Rightarrow$ $\frac{1}{5}$

(b) H : $60^{o} /_{o}$ of the students read Hindi newspaper,

E : $40^{o} /_{o}$ read English newspaper and

$H \cap E :$ $20^{o} /_{o}$ read both Hindi and English newspapers.

$P (H) = \frac{60}{100} = \frac{6}{10} = \frac{3}{5}$

$P (E) = \frac{40}{100} = \frac{4}{10} = \frac{2}{5}$

$P (H \cap E) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$

The probability that she reads English newspaper if she reads Hindi newspaper $= P (E | H)$

$P (E | H) = \frac{P (E \cap H)}{P (H)}$

$P (E | H) = \frac{\frac{1}{5}}{\frac{3}{5}}$

$P (E | H) = \frac{1}{3}$

E : $40^{o} /_{o}$ read English newspaper and

$H \cap E :$ $20^{o} /_{o}$ read both Hindi and English newspapers.

$P (H) = \frac{60}{100} = \frac{6}{10} = \frac{3}{5}$

$P (E) = \frac{40}{100} = \frac{4}{10} = \frac{2}{5}$

$P (H \cap E) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$

the probability that she reads Hindi newspaper if she reads English newspaper $= P (H | E)$

$P (H | E) = \frac{P (H \cap E)}{P (E)}$

$P (H | E) = \frac{\frac{1}{5}}{\frac{2}{5}}$

$P (H | E) = \frac{1}{2}$

Question:17 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) $0$

(B) $\frac{1}{3}$

(D) $\frac{1}{36}$

Answer:

when a pair of dice is rolled, total outcomes $= 6^{2} = 36$

Even prime number $= {2}$

$n (e v e n p r i m e n u m b e r) = 1$

The probability of obtaining an even prime number on each die $= P (E)$

$P (E) = \frac{1}{36}$

Option D is correct.

Question:18 Two events A and B will be independent, if

(A) $A$ and $B$ are mutually exclusive

(B) $P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]$

(D) $P (A) + P (B) = 1$

Answer:

Two events A and B will be independent, if

$P (A \cap B) = P (A) . P (B)$

Or $P (A^{'} \cap B^{'}) = P (A^{'} B^{'}) = P (A^{'}) . P (B^{'}) = (1 - P (A)) . (1 - P (B))$

Option B is correct.

Class 12 Maths chapter 13 solutions - Exercise: 13.3
Page number: 431-433
Total questions: 14

Question:1 An urn contains $5$ red and $5$ black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, $2$ additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Answer:

Black balls = 5

Red balls = 5

Total balls = 10

CASE 1 Let red ball be drawn in first attempt.

$P (d r a w i n g r e d b a l l) = \frac{5}{10} = \frac{1}{2}$

Now two red balls are added in urn .

Now red balls = 7, black balls = 5

Total balls = 12

$P (d r a w i n g r e d b a l l) = \frac{7}{12}$

CASE 2

Let black ball be drawn in first attempt.

$P (d r a w i n g b l a c k b a l l) = \frac{5}{10} = \frac{1}{2}$

Now two black balls are added in urn .

Now red balls = 5, black balls = 7

Total balls = 12

$P (d r a w i n g r e d b a l l) = \frac{5}{12}$

the probability that the second ball is red =

$\Rightarrow$ $\frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12}$

$\Rightarrow$ $\frac{7}{24} + \frac{5}{24}$

$\Rightarrow$ $\frac{12}{24} = \frac{1}{2}$

Question:2 A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn frome bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer:

BAG 1 : Red balls =4 Black balls=4 Total balls = 8

BAG 2 : Red balls = 2 Black balls = 6 Total balls = 8

B1 : selecting bag 1

B2 : selecting bag 2

$P (B 1) = P (B 2) = \frac{1}{2}$

Let R be a event of getting red ball

$P (R | B 1) = P (d r a w i n g r e d b a l l f r o m f i r s t b a g) = \frac{4}{8} = \frac{1}{2}$

$P (R | B 2) = P (d r a w i n g r e d b a l l f r o m s e c o n d b a g) = \frac{2}{8} = \frac{1}{4}$

probability that the ball is drawn from the first bag,

given that it is red is $P (B 1 | R)$ .

Using Baye's theorem, we have

$P (B 1 | R) = \frac{P (B 1) . P (R | B 1)}{P (B 1) . P (R | B 1) + P (B 2) . P (R | B 2)}$

$\Rightarrow$ $P (B 1 | R) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}}$

$\Rightarrow$ $P (B 1 | R) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}}$

$\Rightarrow$ $P (B 1 | R) = \frac{\frac{1}{4}}{\frac{3}{8}}$

$\Rightarrow$ $P (B 1 | R) = \frac{2}{3}$

Question:3 Of the students in a college, it is known that $60^{o} /_{o}$ reside in hostel and $40^{o} /_{o}$ are day scholars (not residing in hostel). Previous year results report that $30^{o} /_{o}$ of all students who reside in hostel attain $A$ grade and $20^{o} /_{o}$ of day scholars attain $A$ grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an $A$ grade, what is the probability that the student is a hostlier?

Answer:

H : reside in hostel

D : day scholars

A : students who attain grade A

$P (H) = \frac{60}{100} = \frac{6}{10} = \frac{3}{5}$

$P (D) = \frac{40}{100} = \frac{4}{10} = \frac{2}{5}$

$P (A | H) = \frac{30}{100} = \frac{3}{10}$

$P (A | D) = \frac{20}{100} = \frac{2}{10} = \frac{1}{5}$

By Bayes theorem :

$P (H | A) = \frac{P (H) . P (A | H)}{P (H) . P (A | H) + P (D) . P (A | D)}$

$\Rightarrow$ $P (H | A) = \frac{\frac{3}{5} \times \frac{3}{10}}{\frac{3}{5} \times \frac{3}{10} + \frac{2}{5} \times \frac{1}{5}}$

$\Rightarrow$ $P (H | A) = \frac{\frac{9}{50}}{\frac{9}{50} + \frac{2}{25}}$

$\Rightarrow$ $P (H | A) = \frac{\frac{9}{50}}{\frac{13}{50}}$

$\Rightarrow$ $P (H | A) = \frac{9}{13}$

Question:4 In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly?

Answer:

A : Student knows answer.

B : Student guess the answer

C : Answer is correct

$P (A) = \frac{3}{4}$ $P (B) = \frac{1}{4}$

$P (C | A) = 1$

$P (C | B) = \frac{1}{4}$

By Bayes theorem :

$P (A | C) = \frac{P (A) . P (C | A)}{P (A) . P (C | A) + P (B) . P (C | B)}$

$\Rightarrow$ $P (A | C) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}}$

$\Rightarrow$ $\frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}}$ $= \frac{\frac{3}{4}}{\frac{13}{16}}$

$\Rightarrow$ $P (A | C) = \frac{12}{13}$

Question:5 A laboratory blood test is $99^{o} /_{o}$ effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for ${0.5}^{o} /_{o}$ of the healthy person tested (i.e. if a healthy person is tested, then, with probability $0.005,$ the test will imply he has the disease). If ${0.1}^{o} /_{o}$ of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?

Answer:

A : Person selected is having the disease

B : Person selected is not having the disease.

C :Blood result is positive.

$P (A) = 0.1 % = \frac{1}{1000} = 0.001$

$P (B) = 1 - P (A) = 1 - 0.001 = 0.999$

$P (C | A) = 99 % = 0.99$

$P (C | B) = 0.5 % = 0.005$

By Bayes theorem :

$P (A | C) = \frac{P (A) . P (C | A)}{P (A) . P (C | A) + P (B) . P (C | B)}$

$\Rightarrow$ $\frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005}$

$\Rightarrow$ $\frac{0.00099}{0.00099 + 0.004995}$

$\Rightarrow$ $\frac{0.00099}{0.005985}$ $= \frac{990}{5985}$

$\Rightarrow$ $\frac{22}{133}$

Question:6 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75^{o} /_{o}$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer:

Given : A : chossing a two headed coin

B : chossing a biased coin

C : chossing a unbiased coin

$P (A) = P (B) = P (C) = \frac{1}{3}$

D : event that coin tossed show head.

$P (D | A) = 1$

Biased coin that comes up heads $75^{o} /_{o}$ of the time.

$P (D | B) = \frac{75}{100} = \frac{3}{4}$

$P (D | C) = \frac{1}{2}$

$P (B | D) = \frac{P (B) . P (D | B)}{P (B) . P (D | B) + P (A) . P (D | A) + P (C) . P (D | C)}$

$\Rightarrow$ $P (B | D) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}$

$\Rightarrow$ $P (B | D) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{4} + \frac{1}{6}}$

$\Rightarrow$ $P (B | D) = \frac{\frac{1}{3}}{\frac{9}{12}}$

$\Rightarrow$ $P (B | D) = \frac{1 \times 12}{3 \times 9}$

$\Rightarrow$ $P (B | D) = \frac{4}{9}$

Question:7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are $0.01, 0.03$ , and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer:

Let A : scooter drivers = 2000

B : car drivers = 4000

C : truck drivers = 6000

Total drivers = 12000

$P (A) = \frac{2000}{12000} = \frac{1}{6} = 0.16$

$P (B) = \frac{4000}{12000} = \frac{1}{3} = 0.33$

$P (C) = \frac{6000}{12000} = \frac{1}{2} = 0.5$

D : the event that person meets with an accident.

$P (D | A) = 0.01$

$P (D | B) = 0.03$

$P (D | C) = 0.15$

$P (A | D) = \frac{P (A) . P (D | A)}{P (B) . P (D | B) + P (A) . P (D | A) + P (C) . P (D | C)}$

$\Rightarrow$ $P (A | D) = \frac{0.16 \times 0.01}{0.16 \times 0.01 + 0.33 \times 0.03 + 0.5 \times 0.15}$

$\Rightarrow$ $P (A | D) = \frac{0.0016}{0.0016 + 0.0099 + 0.075}$

$\Rightarrow$ $P (A | D) = \frac{0.0016}{0.0865}$

$\Rightarrow$ $P (A | D) = 0.019$

Question:8 A factory has two machines $A$ and $B .$ Past record shows that machine $A$ produced $60^{o} /_{o}$ of the items of output and machine B produced $40^{o} /_{o}$ of the items. Further, $2^{o} /_{o}$ of the items produced by machine $A$ and $1^{o} /_{o}$ produced by machine $B$ were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine $B$ ?

Answer:

A : Items produced by machine A $= 60 %$

B : Items produced by machine B $= 40 %$

$P (A) = \frac{60}{100} = \frac{3}{5}$

$P (B) = \frac{40}{100} = \frac{2}{5}$

X : Produced item found to be defective.

$P (X | A) = \frac{2}{100} = \frac{1}{50}$

$P (X | B) = \frac{1}{100}$

$P (B | X) = \frac{P (B) . P (X | B)}{P (B) . P (X | B) + P (A) . P (X | A)}$

$\Rightarrow$ $P (B | X) = \frac{\frac{2}{5} \times \frac{1}{100}}{\frac{2}{5} \times \frac{1}{100} + \frac{3}{5} \times \frac{1}{50}}$

$\Rightarrow$ $P (B | X) = \frac{\frac{1}{250}}{\frac{1}{250} + \frac{3}{250}}$

$\Rightarrow$ $P (B | X) = \frac{\frac{1}{250}}{\frac{4}{250}}$

$\Rightarrow$ $P (B | X) = \frac{1}{4}$

Hence, the probability that defective item was produced by machine $B$ =

$P (B | X) = \frac{1}{4}$ .

Question:9 Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.

Answer:

A: the first groups will win

B: the second groups will win

$P (A) = 0.6$

$P (B) = 0.4$

X: Event of introducing a new product.

Probability of introducing a new product if the first group wins : $P (X | A) = 0.7$

Probability of introducing a new product if the second group wins : $P (X | B) = 0.3$

$P (B | X) = \frac{P (B) . P (X | B)}{P (B) . P (X | B) + P (A) . P (X | A)}$

$\Rightarrow$ $p (B | X) = \frac{0.4 \times 0.3}{0.4 \times 0.3 + 0.6 \times 0.7}$

$\Rightarrow$ $p (B | X) = \frac{0.12}{0.12 + 0.42}$

$\Rightarrow$ $p (B | X) = \frac{0.12}{0.54}$

$\Rightarrow$ $p (B | X) = \frac{12}{54}$

$\Rightarrow$ $p (B | X) = \frac{2}{9}$

Hence, the probability that the new product introduced was by the second group :

$p (B | X) = \frac{2}{9}$

Question:10 Suppose a girl throws a die. If she gets a $5$ or $6$ , she tosses a coin three times and notes the number of heads. If she gets $1, 2, 3$ or $4$ , she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1, 2, 3$ or $4$ with the die?

Answer:

Let, A: Outcome on die is 5 or 6.

B: Outcome on die is 1,2,3,4

$P (A) = \frac{2}{6} = \frac{1}{3}$

$P (B) = \frac{4}{6} = \frac{2}{3}$

X: Event of getting exactly one head.

Probability of getting exactly one head when she tosses a coin three times : $P (X | A) = \frac{3}{8}$

Probability of getting exactly one head when she tosses a coin one time : $P (X | B) = \frac{1}{2}$

$P (B | X) = \frac{P (B) . P (X | B)}{P (B) . P (X | B) + P (A) . P (X | A)}$

$\Rightarrow$ $P (B | X) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{2}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{3}{8}}$

$\Rightarrow$ $P (B | X) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{8}}$

$\Rightarrow$ $P (B | X) = \frac{\frac{1}{3}}{\frac{11}{24}}$

$\Rightarrow$ $P (B | X) = \frac{1 \times 24}{3 \times 11} = \frac{8}{11}$

Hence, the probability that she threw $1, 2, 3$ or $4$ with the die =

$P (B | X) = \frac{8}{11}$

Question:11 A manufacturer has three machine operators $A, B$ and $C .$ The first operator $A$ produces $1^{o} /_{o}$ defective items, where as the other two operators B and C produce $5^{o} /_{o}$ and $7^{o} /_{o}$ defective items respectively. $A$ is on the job for $50^{o} /_{o}$ of the time, $B$ is on the job for $30^{o} /_{o}$ of the time and $C$ is on the job for $20^{o} /_{o}$ of the time. A defective item is produced, what is the probability that it was produced by $A$ ?

Answer:

Let A: time consumed by machine A $= 50 %$

B: time consumed by machine B $= 30 %$

C: time consumed by machine C $= 20 %$

Total drivers = 12000

$P (A) = \frac{50}{100} = \frac{1}{2}$

$P (B) = \frac{30}{100} = \frac{3}{10}$

$P (C) = \frac{20}{100} = \frac{1}{5}$

D: Event of producing defective items

$P (D | A) = \frac{1}{100}$

$P (D | B) = \frac{5}{100}$

$P (D | C) = \frac{7}{100}$

$P (A | D) = \frac{P (A) . P (D | A)}{P (B) . P (D | B) + P (A) . P (D | A) + P (C) . P (D | C)}$

$\Rightarrow$ $P (A | D) = \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{2} \times \frac{1}{100} + \frac{3}{10} \times \frac{5}{100} + \frac{1}{5} \times \frac{7}{100}}$

$\Rightarrow$ $P (A | D) = \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{100} (\frac{1}{2} + \frac{3}{2} + \frac{7}{5})}$

$\Rightarrow$ $P (A | D) = \frac{\frac{1}{2}}{(\frac{17}{5})}$

$\Rightarrow$ $P (A | D) = \frac{5}{34}$

Hence, the probability that defective item was produced by $A$ =

$P (A | D) = \frac{5}{34}$

Question:12 A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer:

Let A : Event of choosing a diamond card.

B : Event of not choosing a diamond card.

$P (A) = \frac{13}{52} = \frac{1}{4}$

$P (B) = \frac{39}{52} = \frac{3}{4}$

X : The lost card.

If lost card is diamond then 12 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 12 diamond cards in $^{12} C_{2}$ ways.

Similarly, two cards are drawn out of 51 cards in $^{51} C_{2}$ ways.

Probablity of getting two diamond cards when one diamond is lost : $P (X | A) = \frac{^{12} C_{2}}{^{51} C_{2}}$

$P (X | A) = \frac{12!}{10! \times 2!} \times \frac{49! \times 2!}{51!}$

$\Rightarrow$ $P (X | A) = \frac{11 \times 12}{50 \times 51}$

$\Rightarrow$ $P (X | A) = \frac{22}{425}$

If lost card is not diamond then 13 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 13 diamond cards in $^{13} C_{2}$ ways.

Similarly, two cards are drawn out of 51 cards in $^{51} C_{2}$ ways.

Probablity of getting two diamond cards when one diamond is not lost : $P (X | B) = \frac{^{13} C_{2}}{^{51} C_{2}}$

$\Rightarrow$ $P (X | B) = \frac{13!}{11! \times 2!} \times \frac{49! \times 2!}{51!}$

$\Rightarrow$ $P (X | B) = \frac{13 \times 12}{50 \times 51}$

$\Rightarrow$ $P (X | B) = \frac{26}{425}$

The probability of the lost card being a diamond : $P (B | X)$

$P (B | X) = \frac{P (B) . P (X | B)}{P (B) . P (X | B) + P (A) . P (X | A)}$

$\Rightarrow$ $P (B | X) = \frac{\frac{1}{4} \times \frac{22}{425}}{\frac{1}{4} \times \frac{22}{425} + \frac{3}{4} \times \frac{26}{425}}$

$\Rightarrow$ $P (B | X) = \frac{\frac{11}{2}}{25}$

$\Rightarrow$ $P (B | X) = \frac{11}{50}$

Hence, the probability of the lost card being a diamond :

$P (B | X) = \frac{11}{50}$

Question:13 Probability that A speaks truth is $\frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is

(A) $\frac{4}{5}$

(B) $\frac{1}{2}$

(D) $\frac{2}{5}$

Answer:

Let A : A speaks truth

B : A speaks false

$P (A) = \frac{4}{5}$

$P (B) = 1 - \frac{4}{5} = \frac{1}{5}$

X : Event that head appears.

A coin is tossed , outcomes are head or tail.

Probability of getting head whether A speaks thruth or not is $\frac{1}{2}$

$P (X | A) = P (X | B) = \frac{1}{2}$

$P (A | X) = \frac{P (A) . P (X | A)}{P (B) . P (X | B) + P (A) . P (X | A)}$

$\Rightarrow$ $P (A | X) = \frac{\frac{4}{5} \times \frac{1}{2}}{\frac{4}{5} \times \frac{1}{2} + \frac{1}{5} \times \frac{1}{2}}$

$\Rightarrow$ $P (A | X) = \frac{\frac{4}{5}}{\frac{4}{5} + \frac{1}{5}}$

$\Rightarrow$ $P (A | X) = \frac{\frac{4}{5}}{\frac{1}{1}}$

$\Rightarrow$ $P (A | X) = \frac{4}{5}$

The probability that actually there was head is $P (A | X) = \frac{4}{5}$

Hence, option A is correct.

Question:14 If $A$ and $B$ are two events such that $A \subset B$ and $P (B) \neq 0,$ then which of the following is correct?

(A) $P (A ∣ B) = \frac{P (B)}{P (A)}$

(B) $P (A ∣ B) < P (A)$

(D) None of these

Answer:

If $A \subset B$ and $P (B) \neq 0,$ then

$\Rightarrow (A \cap B) = A$

Also, $P (A) < P (B)$

$P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (A)}{P (B)}$

We know that $P (B) \leq 1$

$\Rightarrow$ $1 \leq \frac{1}{P (B)}$

$\Rightarrow$ $P (A) \leq \frac{P (A)}{P (B)}$

$\Rightarrow$ $P (A) \leq P (A | B)$

Hence, we can see option C is correct.

Class 12 Maths chapter 13 solutions - Miscellaneous Exercise
Page number: 435-437
Total questions: 13

Question:1 A and B are two events such that $P (A) \neq 0.$ Find $P (B ∣ A),$ if

(i) $A$ is a subset of $B$

(ii) $A \cap B = ϕ$

Answer:

(i) A and B are two events such that $P (A) \neq 0.$

$A \subset B$

$\Rightarrow A \cap B = A$

$P (A \cap B) = P (B \cap A) = P (A)$

$P (B | A) = \frac{P (B \cap A)}{P (A)}$

$\Rightarrow$ $P (B | A) = \frac{P (A)}{P (A)}$

$\Rightarrow$ $P (B | A) = 1$

(ii) A and B are two events such that $P (A) \neq 0.$

$P (A \cap B) = P (B \cap A) = 0$

$P (B | A) = \frac{P (B \cap A)}{P (A)}$

$\Rightarrow$ $P (B | A) = \frac{0}{P (A)}$

$\Rightarrow$ $P (B | A) = 0$

Question:2 A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Answer:

(i) A couple has two children,

sample space $= {(b, b), (g, g), (b, g), (g, b)}$

Let A be both children are males and B is at least one of the children is male.

$(A \cap B) = {(b, b)}$

$P (A \cap B) = \frac{1}{4}$

$P (A) = \frac{1}{4}$

$P (B) = \frac{3}{4}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

(ii) A couple has two children,

sample space $= {(b, b), (g, g), (b, g), (g, b)}$

Let A be both children are females and B be the elder child is a female.

$(A \cap B) = {(g, g)}$

$P (A \cap B) = \frac{1}{4}$

$P (A) = \frac{1}{4}$

$P (B) = \frac{2}{4}$

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

$\Rightarrow$ $P (A | B) = \frac{\frac{1}{4}}{\frac{2}{4}} = \frac{1}{2}$

Question:3 Suppose that $5^{o} /_{o}$ of men and ${0.25}^{o} /_{o}$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Answer:

We have $5^{o} /_{o}$ of men and ${0.25}^{o} /_{o}$ of women have grey hair.

Percentage of people with grey hairs $= (5 + 0.25) % = 5.25 %$

The probability that the selected haired person is male :

$= \frac{5}{5.25} = \frac{20}{21}$

Question:4 Suppose that $90^{o} /_{o}$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Answer:

$90^{o} /_{o}$ of people are right-handed.

$P (r i g h t - h a n d e d) = \frac{9}{10}$

$P (l e f t - h a n d e d) = q = 1 - \frac{9}{10} = \frac{1}{10}$

at most $6$ of a random sample of $10$ people are right-handed.

the probability that more than $6$ of a random sample of $10$ people are right-handed is given by,

$\sum_{T}^{10} ^{10}C_r P^{r} q^{10-r}$ $=\sum_{T}^{10} ^{10}C_r$

${\frac{9}{10}}^{r} . (\frac{1}{10})^{10 - r}$

the probability that at most $6$ of a random sample of $10$ people are right-handed is given by

$=1-\sum_{T}^{10} ^{10}C_r$ . ${\frac{9}{10}}^{r} . (\frac{1}{10})^{10 - r}$

Question:5 If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Answer:

In a leap year, there are 366 days.

In 52 weeks, there are 52 Tuesdays.

The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.

The remaining 2 days can be :

1. Monday and Tuesday

2. Tuesday and Wednesday

3. Wednesday and Thursday

4. Thursday and Friday

5.friday and Saturday

6.saturday and Sunday

7.sunday and Monday

Total cases = 7.

Favorable cases = 2

Probability of having 53 Tuesday in a leap year = P.

$P = \frac{2}{7}$

Question:6 Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

1648018018261

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn

(i) from box A?

(ii) from box B?

(iii) from box C?

Answer:

1648018246557 '

(i) Let R be the event of drawing red marble.

Let $E_{A}, E_{B}, E_{C}$ respectively denote the event of selecting box A, B, C.

Total marbles = 40

Red marbles =15

$P (R) = \frac{15}{40} = \frac{3}{8}$

Probability of drawing red marble from box A is $P (E_{A} | R)$

$P (E_{A} | R) = \frac{P (E_{A} \cap R)}{P (R)}$

$= \frac{\frac{1}{40}}{\frac{3}{8}}$

$= \frac{1}{15}$

(ii) Let R be event of drawing red marble.

Let $E_{A}, E_{B}, E_{C}$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P (R) = \frac{15}{40} = \frac{3}{8}$

Probability of drawing red marble from box B is $P (E_{B} | R)$

$P (E_{B} | R) = \frac{P (E_{B} \cap R)}{P (R)}$

$= \frac{\frac{6}{40}}{\frac{3}{8}}$

$= \frac{2}{5}$

(iii) Let R be event of drawing red marble.

Let $E_{A}, E_{B}, E_{C}$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P (R) = \frac{15}{40} = \frac{3}{8}$

Probability of drawing red marble from box C is $P (E_{C} | R)$

$P (E_{C} | R) = \frac{P (E_{C} \cap R)}{P (R)}$

$= \frac{\frac{8}{40}}{\frac{3}{8}}$

$= \frac{8}{15}$

Question:7 Assume that the chances of a patient having a heart attack is $40^{o} /_{o} .$ It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30^{o} /_{o}$ and prescription of certain drug reduces its chances by $25^{o} /_{o} .$ At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer:

Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted

the drug prescription.

$∴ P (A) = 0.40$

$∴ P (E 1) = P (E 2) = \frac{1}{2}$

$P (A | E 1) = 0.40 \times 0.70 = 0.28$

$P (A | E 2) = 0.40 \times 0.75 = 0.30$

the probability that the patient followed a course of meditation and yoga is $P (E 1, A)$

$P (E 1, A) = \frac{P (E 1) . P (E 1 | A)}{P (E 1) . P (E 1 | A) + P (E 2) . P (E 2 | A)}$

$P (E 1, A) = \frac{0.5 \times 0.28}{0.5 \times 0.28 + 0.5 \times 0.30}$

$= \frac{14}{29}$

Question:8 If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ).

Answer:

Total number of determinant of second order with each element being 0 or 1 is $2^{4} = 16$

The values of determinant is positive in the following cases $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}], [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}]$

Probability is

$= \frac{3}{16}$

Question:9 An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = $0.2$

P(B fails alone) = $0.15$

P(A and B fail) = $0.15$

Evaluate the following probabilities

(i) $P (A f a i l s ∣ B h a s f a i l e d)$

(ii) $P (A f a i l s a l o n e)$

Answer:

(i) Let event in which A fails and B fails be $E_{A}, E_{B}$

$P (E_{A}) = 0.2$

$P (E_{A} a n d E_{B}) = 0.15$

$P (B f a i l s a l o n e) = P (E_{B}) - P (E_{A} a n d E_{B})$

$\Rightarrow 0.15 = P (E_{B}) - 0.15$

$\Rightarrow P (E_{B}) = 0.3$

$P (E_{A} | E_{B}) = \frac{P (E_{A} \cap E_{B})}{P (E_{B})}$

$= \frac{0.15}{0.3} = 0.5$

(ii) Let the 30622event in which A fails and B fails be $E_{A}, E_{B}$

$P (E_{A}) = 0.2$

$P (E_{A} a n d E_{B}) = 0.15$

$P (B f a i l s a l o n e) = P (E_{B}) - P (E_{A} a n d E_{B})$

$\Rightarrow 0.15 = P (E_{B}) - 0.15$

$\Rightarrow P (E_{B}) = 0.3$

$P (A f a i l s a l o n e) = P (E_{A}) - P (E_{A} a n d E_{B})$

$= 0.2 - 0.15 = 0.05$

Question:10 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer:

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.

$P (E 1) = \frac{3}{7}$ and $P (E 2) = \frac{4}{7}$

Let A be the event that ball drawn is red.

When a red ball is transfered from bag 1 to bag 2.

$P (A | E 1) = \frac{5}{10} = \frac{1}{2}$

When a black ball is transfered from bag 1 to bag 2.

$P (A | E 2) = \frac{4}{10} = \frac{2}{5}$

$P (E 2 | A) = \frac{P (E 2) . P (A | E 2)}{P (E 2) . P (A | E 2) + P (E 1) . P (A | E 1)}$

$= \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{4}{7} \times \frac{2}{5} + \frac{3}{7} \times \frac{1}{2}}$

$= \frac{16}{31}$

Question:11 If A and B are two events such that $P (A \neq 0)$ and $P (B ∣ A) = 1,$ then

Choose the correct answer of the following:

(A) $A \subset B$

(B) $B \subset A$

(D) $A = ϕ$

Answer:

A and B are two events such that $P (A \neq 0)$ and $P (B ∣ A) = 1,$

$P (B | A) = \frac{P (B \cap A)}{P (A)}$

$1 = \frac{P (B \cap A)}{P (A)}$

$P (B \cap A) = P (A)$

$\Rightarrow A \subset B$

Option A is correct.

Question:12 If $P (A ∣ B) > P (A)$ , then which of the following is correct :

(A) $P (B ∣ A) < P (B)$

(B) $P (A \cap B) < P (A) . P (B)$

(D) $P (B ∣ A) = P (B)$

Answer:

$P (A ∣ B) > P (A)$

$\Rightarrow \frac{P (A \cap B)}{P (B)} > P (A)$

$\Rightarrow P (A \cap B) > P (A) . P (B)$

$\Rightarrow \frac{P (A \cap B)}{P (A)} > P (B)$

$\Rightarrow P (B | A) > P (B)$

Option C is correct.

Question:13 If A and B are any two events such that $P (A) + P (B) - P (A a n d B) = P (A),$ then

(A) $P (B ∣ A) = 1$

(B) $P (A ∣ B) = 1$

(D) $P (A ∣ B) = 0$

Answer:

$P (A) + P (B) - P (A a n d B) = P (A),$

$P (A) + P (B) - P (A \cap B) = P (A)$

$\Rightarrow P (B) - P (A \cap B) = 0$

$\Rightarrow P (B) = P (A \cap B)$

$P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (B)}{P (B)} = 1$

Option B is correct.

If you are looking for probability class 12 ncert solutions of exercise then they are listed below.

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

Importance of solving NCERT questions for class 12 Math Chapter 13

Here, we cover the topics of Probability, which is important for building a strong understanding of mathematical concepts and developing problem-solving skills. These exercises help in the theoretical knowledge, enhance analytical thinking, and prepare students for exams by providing a wide range of practice problems that align with the syllabus.

As this chapter has 10% weightage in 12th board final exam. NCERT solutions for class 12 maths chapter 13 probability will help you to score good marks in the final exam.
NCERT solutions for Class 12 Maths Chapter 13 make it very easy for you to understand as these are prepared and explained in a detailed manner.
At the end of every chapter, there is an additional exercise called Miscellaneous exercise which is very important for you to develop a grip on the concepts. In NCERT solutions for class 12 maths chapter 13 probability, you will get solutions for miscellaneous exercise too.
These NCERT solutions for Class 12 Maths Chapter 13 PDF download are prepared with different approaches so it will give you new ways of solving the problems.
Experts at Career 360 make NCERT solutions for class 12 maths chapter 13 probability very easy for you to understand and that will help you to score good marks in board exams.

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Happy Reading !!!

Frequently Asked Questions (FAQs)

1. How to solve Bayes' Theorem questions in Class 12 Probability?

To solve Bayes' Theorem questions, apply the formula:

P(A | B) = (P(B | A) x P(A)) / P(B)
Here, P(A | B) is the probability of event A given event B, P(B | A) is the probability of event B given A, and P(A) and P(B) are the probabilities of A and B.

2. What is the formula for Conditional Probability in NCERT Class 12 Maths?

The formula for Conditional Probability is:

P(A | B) = P(A ∩ B) / P(B)

where P(A | B) is the probability of A occurring given that B has occurred.

3. What are the basic concepts of Probability in NCERT Class 12 Maths?

The basic concepts include understanding random experiments, sample spaces, events (simple and compound), probability rules (addition and multiplication rules), and probability distributions.

4. How to solve Bernoulli Trials and Binomial Distribution questions?

To solve these questions, use the Binomial Distribution formula:

P(X = k) = (n choose k) x p^k x (1 - p)^(n - k)

where n is the number of trials, k is the number of successes, and p is the probability of success in each trial.

5. What is the difference between Classical and Conditional Probability in NCERT?

Classical Probability is based on equally likely outcomes (i.e., all outcomes are equally probable), while Conditional Probability is the probability of an event occurring given that another event has already occurred. The key difference is that Classical looks at all possible outcomes equally, while Conditional focuses on a subset of outcomes based on additional information.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 12 Maths Chapter 13 Probability

NCERT Probability Class 12 Questions And Answers PDF Free Download

NCERT Class 12 Maths Chapter 13 Question Answer Probability - Important Formulae

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Importance of solving NCERT questions for class 12 Math Chapter 13

NCERT Exemplar Class 12 Solutions

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