NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3 - Probability

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

$P(B1)=P(B2)=\frac{1}{2}$

Let R be a event of getting red ball

$P(R|B1) = P(drawing\, \, red\, \, ball\, \, from \, first \, \, bag)= \frac{4}{8}=\frac{1}{2}$

$P(R|B2) = P(drawing\, \, red\, \, ball\, \, from \, second \, \, bag)= \frac{2}{8}=\frac{1}{4}$

probability that the ball is drawn from the first bag,

given that it is red is $P(B1|R)$.

Using Baye's theorem, we have

$P(B1|R) = \frac{P(B1).P(R|B1)}{P(B1).P(R|B1)+P(B2).P(R|B2)}$

$P(B1|R) = \frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}$

$P(B1|R) =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}$

$P(B1|R) =\frac{\frac{1}{4}}{\frac{3}{8}}$

$P(B1|R) = \frac{2}{3}$

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 :

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

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

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

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

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

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 :

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

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

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

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

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 :

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

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

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

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

$=\frac{22}{133}$

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.

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

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

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

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

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

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

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

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

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.

$P(D|A)= 0.01$

$P(D|B)= 0.03$

$P(D|C)= 0.15$

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

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

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

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

$P(A|D)= 0.019$

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.

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

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

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

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

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

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

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

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 : $P(X|B)=0.3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $^{12}\textrm{C}_2$ ways.

Similarly, two cards are drawn out of 51 cards in $^{51}\textrm{C}_2$ ways.

Probablity of getting two diamond cards when one diamond is lost : $P(X|A)= \frac{^{12}\textrm{C}_2}{^{51}\textrm{C}_2}$

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

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

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

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 $^{13}\textrm{C}_2$ ways.

Similarly, two cards are drawn out of 51 cards in $^{51}\textrm{C}_2$ ways.

Probablity of getting two diamond cards when one diamond is not lost : $P(X|B)= \frac{^{13}\textrm{C}_2}{^{51}\textrm{C}_2}$

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

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

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

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

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

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

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

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

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) $\frac{1}{2}$

C) $\frac{1}{5}$

(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 $\frac{1}{2}$

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

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

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

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

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

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

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) $P(A\mid B)=\frac{P(B)}{P(A)}$

(B) $P(A\mid B)< P(A)$

(C) $P(A\mid B)\geq P(A)$

(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 $P(B)\leq 1$

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

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

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

Hence, we can see option C is correct.