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

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:01 PM IST

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.

## Probability Excercise: 30.1

Probability exercise 30.1 question 1
$\frac{4}{7}$
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

$\frac{1}{2}, \frac{1}{3}$
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)
1. 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

$\frac{1}{15}$
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

$\frac{1}{2}$
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

$\frac{1}{6}$
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

$=\frac{16}{25}$
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

\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

$\frac{5}{6},\frac{1}{2}$
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

$\frac{1}{3},\frac{1}{2}$
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)
1. 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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## RD Sharma Chapter-wise Solutions

1. What is the probability formula?

Probability = (number of a favourable outcome)/ (total number of outcomes)

P= n(E) / n(S) where P is the probability, E is the event and S is the sample space.

2. What are the three types of probability?

There are commonly three perspectives on probability-  classical,  empirical,  subjective.

3. What is the probability of getting an even number?

The probability of getting an even number on the first side when a pair of dice is thrown once is 1/2.

4. What is simple probability?

Simple probability is the calculation of an outcome or the chance of an event ever happening.

Yes, RD Shrama class 12 probability is of the latest version which means students can find the latest questions easily.

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