RD Sharma Class 12 Exercise 30.1 Probability Solutions Maths-Download PDF Online

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The RD Sharma textbook solutions are famous in the entire country for their simple understanding and elaboration for the best knowledge for students. The Class 12 RD Sharma chapter 30 exercise 30.1 solution covers the content of the chapter Probability. RD Sharma solutions This chapter is thought to be complex for students' understanding, but once the basic knowledge is acquired from the right source, it might seem a lot more interesting to solve. The RD Sharma class 12th exercise 30.1 is useful in clearing the basics of the chapter and gives insights into how easily the questions can be solved.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.1
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.1

Probability exercise 30.1 question 1
Answer:
$\frac{4}{7}$
Hint:
Use Formula
$\text { Probability }=\frac{ \text { No. of events }}{\text { Total no. of events }}$
Given,
ten cards in a box.
So, let
Total no. of cards,
S = { 1,2,3,4,5,6,7,8,9,10 }
Consider the given events,
A = Total even no. appears on the card
B = No. from cards which is always more than 3
Therefore,
A = { 2,4,6,8,10 }
B = { 4,5,6,7,8,9,10 }
Now,
A $\cap$ B = { 4,6,8,10 }
Therefore,
Probability that card drawn in even no.
$\begin{aligned} &P(A\mid B)=\frac{ \text { No. of elements of }A\cap B}{\text { Total no. of cards more than 3 }}\\ &\qquad \qquad=\frac{4}{7} \end{aligned}$

Probability exercise 30.1 question 2

Answer:
$\frac{1}{2}, \frac{1}{3}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { No. of event happened }}{ \text { Total no. of events }}$
As per question,
Let , S = Elements of all certain possibilities
= { (B₁B₂), (B₁G₂), (G₁G₂), (G₁B₂) }
Now, A = Event where both child is girl
= { (G1 , G2) }
B = Event where youngest child is a girl
= { (B₁G₂), (G₁G₂) }
C = At Least one child is a girl
= { (B₁G₂), (G₁G₂), (G₁B₂) }
Now,
A ∩ B = (G₁G₂)
A ∩ C = (G₁G₂)

Now,

Required probability
$\begin{aligned} &=\frac{n(A \cap B)}{n(B)}\\ &=\frac{1}{2} \end{aligned}$
2. Required probability
$\begin{aligned} &=\frac{n(A \cap B)}{n(C)}\\ &=\frac{1}{3} \end{aligned}$

Probability exercise 30.1 question 3

Answer:
$\frac{1}{15}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { Required event }}{ \text { Total no. of event }}$
Given, an event where two numbers appear on throwing two dice differently.
So, as per the question
Let, A = No on dice where two values are different
= { (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,6) }
Similarly,
B= Event where sum of the no. is 4
= (1,3), (2,2), (3,1)
Now,
For event where no. is different and also its sum is 4
$\begin{aligned} &=A \cap B\\ &=\left \{ \left \{ 1,3 \right \}, \left \{ 3,1 \right \} \right \} \end{aligned}$
Thus,
$\begin{aligned} &\text { Probability }=\frac{n(A\cap B)}{ nA }\\ &=\frac{2}{30}\\ &=\frac{1}{15} \end{aligned}$

Probability exercise 30.1 question 4

Answer:
$\frac{1}{2}$
Hint:
$\text { Probability }=\frac{\text { No. of elements required }}{ \text { Total no. of elements }}$
Given, two events
Let, B = Event of getting head on toss third
= { HHH, HTH, THH, TTH }
Similarly, A = Event of getting head on first two tosses
= { HHH, HHT }
Now,
A∩B = { HHH }
Therefore,
Required probability
$\begin{aligned} &=\frac{n(A\cap B)}{n(A)}\\ &=\frac{1}{2} \end{aligned}$

Probability exercise 30.1 question 5

Answer:
$\frac{1}{6}$
Hint:
Find all the possibilities of getting 4 on third toss.
Given, that 6 and 5 appear respectively on the first two tosses.
So, let
A = Event for getting 4 on third toss
= { (1,1,4), (1,2,4), (1,3,4), (1,4,4), (1,5,4), (1,6,4),
(2,1,4), (2,2,4), (2,3,4), (2,4,4), (2,5,4), (2,6,4),
(3,1,4), (3,2,4), (3,3,4), (3,4,4), (3,5,4), (3,6,4),
(4,1,4), (4,2,4), (4,3,4), (4,4,4), (4,5,4), (4,6,4),
(5,1,4), (5,2,4), (5,3,4), (5,4,4), (5,5,4), (5,6,4),
(6,1,4), (6,2,4), (6,3,4), (6,4,4), (6,5,4), (6,6,4) }
Now, let
B = Event of getting 6 on first and 5 on second
= (6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6),
Now,
A ∩ B = (6,5,4)
Therefore,
$\begin{aligned} &\text { Probability }=\frac{n(A\cap B)}{ n(B)}\\ &=\frac{1}{6} \end{aligned}$

Probability exercise 30.1 question 6

Answer:
$=\frac{16}{25}$
Hint:
Just put values and simplify
Given, P (B) = 0.5
$\begin{aligned} &P(A\cap B)=0.32\\ &P(A\mid B)=\frac{P(A\cap B)}{P(B)}\\ &=\frac{0.32}{0.5}\\ &=\frac{32}{50}\\ &=\frac{16}{25} \end{aligned}$

Probability exercise 30.1 question 7

Answer:
$\begin{aligned} &0.2, \frac{2}{3} \end{aligned}$
Hint:
Just simplify to find P (A|B)
Given, P(A) = 0.4
P(B) = 0.3
Now
$\begin{aligned} &P(A\mid B)=\frac{n(A\cap B)}{ n(A)}\\ &0.5=\frac{n(A\cap B)}{ 0.4 }\\ &P(A\cap B)=0.20 \end{aligned}$
Therefore,
$\begin{aligned} &P(A\mid B)=\frac{n(A\cap B)}{ n(B)}\\ &=\frac{0.2}{ 0.3 }\\ &=\frac{2}{3} \end{aligned}$

Probability exercise 30.1 question 8

Answer:
$\frac{5}{6},\frac{1}{2}$
Hint:
Use Formula
$\begin{aligned} &P(A\cup B)=P(A)+P(B)-P(A\cap B) \end{aligned}$ +
Given,
$\begin{aligned} &P(A)=\frac{1}{3}\\ &P(B)=\frac{1}{5}\\ &P(A\cup B)=\frac{11}{30}\\ \end{aligned}$
We know that,
$\begin{aligned} &P(A\cup B)=P(A)+P(B)-P(A\cap B) \end{aligned}$
$\begin{aligned} &\frac{11}{30}=\frac{1}{3}+\frac{1}{5}-P(A\cap B)\\ &P(A\cap B)=\frac{8}{15}-\frac{11}{30}\\ &P(A\cap B)=\frac{16-11}{30}\\ &\qquad \qquad =\frac{5}{30}\\ &P(A\cap B)=\frac{1}{6} \end{aligned}$
Now for,
$\begin{aligned} &P(A\mid B)=\frac{P(A\cap B)}{P(A)}\\ &=\frac{\frac{1}{6}}{\frac{1}{5}}\\ &=\frac{5}{6}\\ &P(B\mid A)=\frac{P(A\cap B)}{P(B)}\\ &=\frac{\frac{1}{6}}{\frac{1}{3}}\\ &=\frac{3}{6}\\ &=\frac{1}{2} \end{aligned}$

Probability exercise 30.1 question 9

Answer:
$\frac{1}{3},\frac{1}{2}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { No. of event happened }}{ \text { Total no. of events }}$
Given, conditions
Let, the whole possibility
S = { (M₁M₂), (M₁F₂), (F₁F₂), (F₁M₂) }
Let , A = Both child are female
= (F₁F₂)
B = Elder child is female
= { (F₁M₂), (F₁F₂) }
C = At Least one child is a male
= { (M₁F₁), (F₁M₂), (M₁M₂) }
D = Both children are male
= (M₁M₂)
D ∩ C = (M₁M₂)
A ∩ B = (F₁F₂)

Required probability

$P(D\mid C)=\frac{n(D\cap C)}{ n(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
2. Required probability
$P(A\mid B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$

The RD Sharma class 12 solution of Probability exercise 30.1 consists of a total of 9 questions which are brief enough to make you understand the basics of the chapter and prepare you to face more tough questions in further exercises. The RD Sharma class 12th exercise 30.1 covers all the concepts of this chapter that are mentioned below-

Conditional Probability
Mutually exclusive event
Union and Intersection
Equally Likely events

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