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

Lovekush kumar sainiUpdated on 25 Jan 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-

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.2
RD Sharma Chapter-wise Solutions

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

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.2

Probability exercise 30.2 question 1

Answer:
$\frac{1}{221}$
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

Answer:
$\frac{1}{270725}$
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 2^nd draw.
C = An ace in the 3^rd draw.
D = An ace in the 4^th 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

Answer:
$\frac{11}{50}$
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

Answer:
$\frac{11}{850}$
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 1^st draw
B = An ace in the 2^nd draw
C = Getting an ace in the 3^rd 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)

Answer:
$\frac{1}{221}$
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 1^st draw
B= A king in 2^nd 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)

Answer:
$\frac{4}{663}$
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 1^st draw
B= An ace in the 2^nd 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)

Answer:
$\frac{25}{204}$
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 1^st throw
D=A red card in the 2^nd 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

Answer:
$\frac{5}{19}$
Hint:
Just take events and apply probability.
Given,
A=An even number in the 1^st draw
B= An odd number in the 2^nd 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

Answer:
$\begin{aligned} &\frac{15}{22} \end{aligned}$
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 1^st draw.
B=A white or red ball in the 2^nd 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

Answer:
$\frac{8}{65}$
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 1^st draw
B=A white or black ball in the 2^nd draw
C=A white or black ball in the 3^rd 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

Answer:
$\frac{13}{204}$
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 1^stdraw
B=A diamond in the 2^nddraw
$\begin{aligned} &P(A)=\frac{13}{52}=\frac{1}{4}\\ &P\left ( \frac{B}{A} \right )=\frac{13}{51}\\ &\text { Required Probability }=P(A\cap B)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\\ &=\frac{1}{4}\times \frac{13}{51}\\ &=\frac{13}{204} \end{aligned}$

Probability exercise 30.2 question 11

Answer:
$\frac{3}{7}$
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 1^st draw
B=A black ball in the 2^nd 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

Answer:
$\begin{aligned} &\frac{2}{5525} \end{aligned}$
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 1^st draw
B=A king in the 2^nd 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

Answer:
$\begin{aligned} &\frac{44}{91} \end{aligned}$
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 1^st draw
B=A good orange in the 2^nd draw
C=A good orange in the 3^rd 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

Answer:
$\begin{aligned} &\frac{1}{24} \end{aligned}$
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 1^st draw
B=A black ball in the 2^nd draw
C=A red ball in the 3^rd 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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