# RD Sharma Class 12 Exercise 30.2 Probability Solutions Maths-Download PDF Online

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

The RD Sharma class 12 solution Probability exercise 30.2 is a tricky chapter to be solved by any student. The RD Sharma class 12th exercise 30.2 starts with the explanation of the basic concepts of the chapter and then moves on to introducing the challenging questions to prepare the students in every way out to score high in the maths exam. The Class 12 RD Sharma chapter 30 exercise 30.2 solution consists of a total of 16 questions to give a brief of the essential concepts of the chapter that are mentioned below-

• The axiomatic approach of Probability

• The classical theory of Probability

• Conditional Probability

• Permutation and Combination

• With replacement and without replacement type questions

• Union and Intersection

• Question based on Cards and balls

## Probability Excercise: 30.2

Probability exercise 30.2 question 1

$\frac{1}{221}$
Hint:
Use formula of probability, where
$P(A\cap B)=P(A)\times P\left ( \frac{B}{A} \right )$
Given, a pack of 52 cards.
Solution: So Let, A = A king in first trial
B = A king in second trial.
Now, Total cards are 52.
\begin{aligned} &P(A)=\frac{ \text { No. of elements of A }}{\text { Total no. of elements }}\\ &=\frac{4}{52}=\frac{1}{13} \end{aligned}
Now
\begin{aligned} &P\left ( \frac{B}{A} \right )=\text { A consecutive king without replacement after first trial.}\\ &=\frac{\text { No. of kings in cards }}{\text { Total number of cards after excluding one king }}\\ &=\frac{3}{51} \qquad \text { (After trail cards will be and there will be kings) }\\ &=\frac{1}{17} \end{aligned}
Now,
\begin{aligned} &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{13}\times \frac{1}{17}\\ &=\frac{1}{221} \end{aligned}

Probability exercise 30.2 question 2

$\frac{1}{270725}$
Hint:
$\text { Probability }=\frac{\text { No. of favourable outcomes }}{ \text { Total no. of outcomes }}$
Solution:
Let, A = An ace in the first draw.
B = An ace in the 2nd draw.
C = An ace in the 3rd draw.
D = An ace in the 4th draw.
\begin{aligned} &\text { Probability of A }=\frac{4}{25}\\ &=\frac{1}{13}\\ &P\left ( \frac{B}{A} \right )=\frac{3}{51}=\frac{1}{17}\\ &P\left ( \frac{C}{A\cap B} \right )=\frac{2}{50}=\frac{1}{25}\\ &P\left ( \frac{D}{A\cap B\cap C} \right )=\frac{1}{49} \end{aligned}
\begin{aligned} &\text { Required Probability }=P\left ( A\cap B\cap C\cap D \right )\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{A\cap B} \right )\times P\left ( \frac{D}{A\cap B\cap C} \right )\\ &=\frac{1}{13}\times \frac{1}{17}\times \frac{1}{25}\times \frac{1}{49}\\ &=\frac{1}{270725} \end{aligned}

Probability exercise 30.2 question 3

### Answer:$\frac{7}{22}$Hint:\begin{aligned} &\text { Probability }=\frac{\text { No. of favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}Let, A = White ball in first draw.B = White ball in second draw.Solution:Now,\begin{aligned} &P(A)=\frac{7}{12}\\ &P\left ( \frac{B}{A} \right )=\frac{6}{11}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{7}{12}\times \frac{6}{11}\\ &=\frac{42}{132}\\ &=\frac{7}{22} \end{aligned}

Probability exercise 30.2 question 4

$\frac{11}{50}$
Hint:
Consider events for tickets and use probability formula
Given, 12 even number between 1 to 25
Solution:
Let,
A = An even number ticket in the first draw
B = An even number ticket in the second draw
Now,
\begin{aligned} &P(A)=\frac{12}{25}\\ &P\left (\frac{B}{A} \right )=\frac{11}{25}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{15}{25}\times \frac{11}{24}\\ &=\frac{11}{50} \end{aligned}

Probability exercise 30.2 question 5

$\frac{11}{850}$
Hint:
To find probability that each time it is card of a spade, calculate the events and take ratio with total events
Let, A = An ace in the 1st draw
B = An ace in the 2nd draw
C = Getting an ace in the 3rd draw.
Now,
\begin{aligned} &P(A)=\frac{13}{52}=\frac{1}{4}\\ &P\left ( \frac{B}{A} \right )=\frac{12}{51}=\frac{4}{17}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{11}{50}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times \frac{C}{B\cap A}\\ &=\frac{1}{4}\times \frac{4}{17}\times \frac{11}{50}\\ &=\frac{11}{850} \end{aligned}

Probability exercise 30.2 question 6(i)

$\frac{1}{221}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given events
Let A=A king in 1st draw
B= A king in 2nd draw
Solution:
\begin{aligned} &P(A)=\frac{4}{52}=\frac{1}{13}\\ &P\left ( \frac{B}{A} \right )=\frac{3}{51}=\frac{1}{17}\\ &\text{ Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{13}\times \frac{1}{17}\\ &=\frac{1}{221} \end{aligned}

Probability exercise 30.2 question 6(ii)

$\frac{4}{663}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given Events
Let A= A king in 1st draw
B= An ace in the 2nd draw
Solution:
\begin{aligned} &P(A)=\frac{4}{52}=\frac{1}{13}\\ &P\left ( \frac{B}{A} \right )=\frac{4}{51}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{13}\times \frac{4}{51}\\ &=\frac{4}{663} \end{aligned}

Probability exercise 30.2 question 6(iii)

$\frac{25}{204}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given, events as follows
Let C=A heart in the 1st throw
D=A red card in the 2nd throw
Now,
\begin{aligned} &P(C)=\frac{13}{52}=\frac{1}{4}\\ &P\left ( \frac{C}{D} \right )=\frac{25}{51}\\ &\text { Required Probability }=P(C\cap D)\\ &=P(C)\times P\left ( \frac{C}{D} \right )\\ &=\frac{1}{4}\times \frac{25}{51}\\ &=\frac{25}{204} \end{aligned}

Probability exercise 30.2 question 7

$\frac{5}{19}$
Hint:
Just take events and apply probability.
Given,
A=An even number in the 1st draw
B= An odd number in the 2nd draw
Now,
\begin{aligned} &P(A)=\frac{10}{20}=\frac{1}{2}\\ &P\left ( \frac{B}{A} \right )=\frac{10}{19}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{2}\times \frac{10}{19}\\ &=\frac{5}{19} \end{aligned}

Probability exercise 30.2 question 8

\begin{aligned} &\frac{15}{22} \end{aligned}
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given events
A=A white or red ball in the 1st draw.
B=A white or red ball in the 2nd draw.
Now,
\begin{aligned} &P(A)=\frac{7}{12}\\ &P\left ( \frac{B}{A} \right )=\frac{6}{11}\\ &P(A\cap B)=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{7}{12}\times \frac{6}{11}\\ &=\frac{7}{22}\\ &\text { Required Probability }=1-P(A\cap B)\\ &=1-\frac{7}{22}\\ &=\frac{15}{22} \end{aligned}

Probability exercise 30.2 question 9

$\frac{8}{65}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given, events
Let A=A white or black ball in the 1st draw
B=A white or black ball in the 2nd draw
C=A white or black ball in the 3rd draw
Now,
\begin{aligned} &P(A)=\frac{8}{15}\\ &P\left ( \frac{B}{A} \right )=\frac{7}{14}=\frac{1}{2}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{6}{13}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right )\\ &=\frac{8}{15}\times \frac{1}{2}\times \frac{6}{13}\\ &=\frac{8}{65} \end{aligned}

Probability exercise 30.2 question 10

$\frac{13}{204}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given, A=A heart in the 1st draw
B=A diamond in the 2nd draw
\begin{aligned} &P(A)=\frac{13}{52}=\frac{1}{4}\\ &P\left ( \frac{B}{A} \right )=\frac{13}{51}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{4}\times \frac{13}{51}\\ &=\frac{13}{204} \end{aligned}

Probability exercise 30.2 question 11

$\frac{3}{7}$
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given, events
A=A black ball in the 1st draw
B=A black ball in the 2nd draw
Now,
\begin{aligned} &P(A)=\frac{10}{15}=\frac{2}{3}\\ &P\left ( \frac{B}{A} \right )=\frac{9}{14}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{2}{3}\times \frac{9}{14}\\ &=\frac{3}{7} \end{aligned}

Probability exercise 30.2 question 12

\begin{aligned} &\frac{2}{5525} \end{aligned}
Hint:
\begin{aligned} &\text { Probability }=P(A\cap B\cap C)=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right ) \end{aligned}
Given, events
A=A king in the 1st draw
B=A king in the 2nd draw
\begin{aligned} &P(A)=\frac{4}{52}=\frac{1}{13}\\ &P\left ( \frac{B}{A} \right )=\frac{3}{51}=\frac{1}{17}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{4}{50}=\frac{2}{25}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right )\\ &=\frac{1}{13}\times \frac{1}{17}\times \frac{2}{25}\\ &=\frac{2}{5525} \end{aligned}

Probability exercise 30.2 question 13

\begin{aligned} &\frac{44}{91} \end{aligned}
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given,
A=A good orange in the 1st draw
B=A good orange in the 2nd draw
C=A good orange in the 3rd draw
Now,
\begin{aligned} &P(A)=\frac{12}{15}=\frac{4}{5}\\ &P\left ( \frac{B}{A} \right )=\frac{11}{14}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{10}{13}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right )\\ &=\frac{4}{5}\times \frac{11}{14}\times \frac{10}{13}\\ &=\frac{440}{910}=\frac{44}{91} \end{aligned}

Probability exercise 30.2 question 14

\begin{aligned} &\frac{1}{24} \end{aligned}
Hint:
\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}
Given,
A=A white ball in the 1st draw
B=A black ball in the 2nd draw
C=A red ball in the 3rd draw
\begin{aligned} &P(A)=\frac{4}{16}=\frac{1}{4}\\ &P\left ( \frac{B}{A} \right )=\frac{7}{15}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{5}{14}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right )\\ &=\frac{1}{4}\times \frac{7}{15}\times \frac{5}{14}\\ &=\frac{1}{24} \end{aligned}

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## RD Sharma Chapter-wise Solutions

1. What is the cost of the RD Sharma class 12 chapter 30 solution?

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2. Is Probability an important chapter for the exam?

Yes, every chapter is important in its own way for the preparation of the board exam. So practice it thoroughly to score high.

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