Anand Classes provides Probability NCERT Solutions Exercise 13.1 Class 12 Chapter-13 Math Notes PDF Free Download (Set-2) provides students with well-structured and exam-oriented solutions designed to strengthen their understanding of fundamental probability concepts. These notes offer clear explanations, step-by-step methods, and accurate solutions following the latest NCERT guidelines, making them ideal for CBSE Class 12 board exam preparation. With comprehensive examples and simplified approaches, this set helps students build confidence and score higher marks. Click the print button to download study material and notes.
Access NCERT Solutions for Probability Exercise 13.1 of Class 12 Math Chapter 13
NCERT Question.9 : Mother, father and son line up at random for a family picture.
Determine $P(E \mid F)$ where
$E$ : son on one end
$F$ : father in the middle
Solution
We need to find the probability that the son is on one end, given that the father is in the middle.
So, let E: Son on one end
F: Father in middle
Event
$$E=\{(M,F,S),(F,M,S),(S,M,F),(S,F,M)\}$$
$$P(E)=\frac{4}{6}=\frac{2}{3}$$
Event
$$F=\{(M,F,S),(S,F,M)\}$$
$$P(F)=\frac{2}{6}=\frac{1}{3}$$
Intersection
$$E \cap F=\{(M,F,S),(S,F,M)\}$$
$$P(E \cap F)=\frac{2}{6}=\frac{1}{3}$$
Thus
$$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$$
Final Result
$$\boxed{P(E \mid F)=1}$$
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NCERT Question 10: A black and a red dice are rolled
(a) Find the Conditional probability of obtaining a sum greater than 9, given that the black die resulted in 5
(b) Find the Conditional probability of obtaining a sum of 8, given that the red die resulted in a number less than 4
Solution :
(a) Conditional probability of obtaining a sum greater than 9, given that the black die resulted in 5
Let $E$: Sum of numbers greater than $9$
Let $F$: Black die shows $5$
We need to find $P(E \mid F)$.
Event $E$
$$E = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$$
$$P(E) = \frac{6}{36}$$
Event $F$
$$F = \{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$$
$$P(F) = \frac{6}{36}$$
Intersection $E \cap F$
$$E \cap F = \{(5,5), (5,6)\}$$
$$P(E \cap F) = \frac{2}{36}$$
Conditional Probability
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{\dfrac{2}{36}}{\dfrac{6}{36}} = \frac{1}{3}$$
Required probability is $\dfrac{1}{3}$.
(b) Conditional probability of obtaining a sum of 8, given that the red die resulted in a number less than 4
Let $E$: Sum of numbers is $8$
Let $F$: Number on red die is less than $4$
We need to find $P(E \mid F)$.
Event $E$
$$E = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$$
$$P(E) = \frac{5}{36}$$
Event $F$
Red die can be $1,2,3$:
$$F = \{(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),\\[1em] (1,2),(2,2),(3,2),(4,2),(5,2),(6,2),
\\[1em] (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\}$$
$$P(F) = \frac{18}{36}$$
Intersection $E \cap F$
$$E \cap F = \{(5,3), (6,2)\}$$
$$P(E \cap F) = \frac{2}{36}$$
Conditional Probability
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{\dfrac{2}{36}}{\dfrac{18}{36}} = \frac{1}{9}$$
Required probability is $\dfrac{1}{9}$.
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NCERT Question.11 : A fair die is rolled.
Consider events $E=\{1,3,5\}$, $F=\{2,3\}$ and $G=\{2,3,4,5\}$. Find
(i) $P(E\mid F)$ and $P(F\mid E)$
(ii) $P(E\mid G)$ and $P(G\mid E)$
(iii) $P\bigl((E\cup F)\mid G\bigr)$ and $P\bigl((E\cap F)\mid G\bigr)$
Solution
A fair die is rolled. Total outcomes $=6$.
(i)
$$(E\cap F)=\{3\}$$
$$P(E\cap F)=\frac{1}{6}$$
$$P(F)=\frac{|\{2,3\}|}{6}=\frac{2}{6}=\frac{1}{3}$$
$$P(E)=\frac{|\{1,3,5\}|}{6}=\frac{3}{6}=\frac{1}{2}$$
Now
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}$$
and
$$P(F\mid E)=\frac{P(F\cap E)}{P(E)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}$$
(ii)
$$(E\cap G)=\{3,5\}$$
$$P(E\cap G)=\frac{2}{6}=\frac{1}{3}$$
$$P(G)=\frac{|\{2,3,4,5\}|}{6}=\frac{4}{6}=\frac{2}{3}$$
So
$$P(E\mid G)=\frac{P(E\cap G)}{P(G)}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$$
and
$$P(G\mid E)=\frac{P(G\cap E)}{P(E)}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$$
(iii)
Let $A=(E\cup F)=\{1,2,3,5\}$ so $P(A)=\frac{4}{6}=\frac{2}{3}$.
$$(A\cap G)={2,3,5}\quad\Rightarrow\quad P(A\cap G)=\frac{3}{6}=\frac{1}{2}$$
Hence
$$P\bigl((E\cup F)\mid G\bigr)=\frac{P(A\cap G)}{P(G)}=\frac{\tfrac{1}{2}}{\tfrac{2}{3}}=\frac{3}{4}$$
Let $B=(E\cap F)={3}$ so $P(B)=\frac{1}{6}$.
$$(B\cap G)={3}\quad\Rightarrow\quad P(B\cap G)=\frac{1}{6}$$
Hence
$$P\bigl((E\cap F)\mid G\bigr)=\frac{P(B\cap G)}{P(G)}=\frac{\tfrac{1}{6}}{\tfrac{2}{3}}=\frac{1}{4}$$
Final Result
$$\boxed{P(E\mid F)=\tfrac{1}{2}, P(F\mid E)=\tfrac{1}{3}, P(E\mid G)=\tfrac{1}{2}, \\[1em]P(G\mid E)=\tfrac{2}{3}, P\bigl((E\cup F)\mid G\bigr)=\tfrac{3}{4}, P\bigl((E\cap F)\mid G\bigr)=\tfrac{1}{4}}$$
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NCERT Question.12 : Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl
(ii) at least one is a girl
Solution
Sample space:
$$S=\{(g,g),(g,b),(b,g),(b,b)\}$$
(i) Youngest child is a girl
Condition: youngest is a girl
So the reduced sample space is:
$$\{(g,g),(b,g)\}$$
Event $E$: both children are girls
$$E=\{(g,g)\}$$
Thus
$$P(E\mid \text{youngest is girl})=\frac{1}{2}$$
(ii) At least one child is a girl
Event $E$: both children are girls
$$E=\{(g,g)\}\quad\Rightarrow\quad P(E)=\frac{1}{4}$$
Event $F$: at least one child is a girl
$$F=\{(g,g),(g,b),(b,g)\}\quad\Rightarrow\quad P(F)=\frac{3}{4}$$
Intersection:
$$E\cap F=\{(g,g)\}$$
$$P(E\cap F)=\frac{1}{4}$$
Now
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac14}{\frac34}=\frac13$$
Final Result
$$\boxed{P(\text{both girls}\mid \text{youngest is girl})=\tfrac12,\\[1em] P(\text{both girls}\mid \text{at least one girl})=\tfrac13}$$
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NCERT Question.13 : An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple-choice questions and 400 difficult multiple-choice questions. If a question is selected at random from the bank, find the probability that it will be an easy question given that it is a multiple-choice question.
Solution
Let
- $E$: easy questions
- $D$: difficult questions
- $M$: multiple-choice questions
- $T$: True/False questions
Given:
$$(E\cap T)=300,\quad (D\cap T)=200,\quad (E\cap M)=500,\quad (D\cap M)=400$$
Total questions:
$$300+200+500+400=1400$$
Event $(E\cap M)$: easy multiple-choice questions
$$P(E\cap M)=\frac{500}{1400}=\frac{5}{14}$$
Event $M$: all multiple-choice questions
$$P(M)=\frac{500+400}{1400}=\frac{900}{1400}=\frac{9}{14}$$
Conditional probability:
$$P(E\mid M)=\frac{P(E\cap M)}{P(M)}=\frac{\tfrac{5}{14}}{\tfrac{9}{14}}=\frac{5}{9}$$
Final Result
$$\boxed{\frac{5}{9}}$$
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NCERT Question.14 : Given that the two numbers appearing on throwing two dice are different, find the probability of the event “the sum of numbers on the dice is 4”.
Solution
Let
- $E$: sum of numbers is $4$
- $F$: numbers on the two dice are different
Event $E$:
$$E=\{(1,3),(3,1),(2,2)\}$$
$$P(E)=\frac{3}{36}=\frac{1}{12}$$
Event $F$: numbers are different
There are $30$ such outcomes out of $36$ total.
$$P(F)=\frac{30}{36}=\frac{5}{6}$$
Intersection:
$$E\cap F=\{(1,3),(3,1)\}$$
$$P(E\cap F)=\frac{2}{36}=\frac{1}{18}$$
Conditional probability:
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15}$$
Final Result
$$\boxed{\frac{1}{15}}$$
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NCERT Question.15 : Consider the experiment of throwing a die. If a multiple of 3 comes up, throw the die again; if any other number comes, toss a coin. Find the conditional probability of the event “the coin shows a tail”, given that “at least one die shows a 3”.
Solution
Let
- $E$: the coin shows a tail
- $F$: at least one die shows a $3$
Event $E$:
A coin is tossed only when the first die shows a non-multiple of $3$, i.e., $\{1,2,4,5\}$.
A tail can appear in the following outcomes:
$$\{(1,T); (2,T); (4,T); (5,T)\}$$
Each die outcome has probability $\dfrac{1}{6}$ and the coin shows tail with probability $\dfrac{1}{2}$:
$$P(E)=4 \times \left(\frac{1}{6}\times \frac{1}{2}\right)=\frac{4}{12}=\frac{1}{3}$$
Event $F$: at least one die shows a $3$
This happens either when the first die is $3$ (then the die is rolled again), or when the second die (in case first was $3$ or $6$) shows $3$.
Possible outcomes in $F$:
$$\{(3,1); (3,2); (3,3); (3,4); (3,5); (3,6); (6,3)\}$$
Thus,
$$P(F)=\frac{7}{36}$$
Intersection $E\cap F$:
A coin is tossed only when the first die is not a multiple of $3$.
But $F$ requires at least one die showing $3$.
So coin toss cannot occur under event $F$.
Thus,
$$E\cap F=\varnothing$$
$$P(E\cap F)=0$$
Conditional probability:
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{\frac{7}{36}}=0$$
Final Result
$$\boxed{0}$$
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NCERT Question.16 : If $P(A)=\frac12$ and $P(B)=0$, then $P(A\mid B)$ is:
(A) 0 (B) $\frac12$ (C) not defined (D) 1
Solution
Given:
$$P(A)=\frac12, P(B)=0$$
Since $P(B)=0$, event $B$ is the null set:
$$B=\varnothing$$
Thus,
$$A\cap B=\varnothing$$
$$P(A\cap B)=0$$
Conditional probability:
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0}{0}$$
The value $\dfrac{0}{0}$ is not defined.
Final Result
$$\boxed{\text{Not defined}}$$
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NCERT Question.17 : If $A$ and $B$ are events such that
$P(A\mid B)=P(B\mid A)$, then
(A) $A\subset B$ but $A\neq B$
(B) $A=B$
(C) $A\cap B=\varnothing$
(D) $P(A)=P(B)$
Solution
Given:
$$P(A\mid B)=P(B\mid A)$$
Using the definition of conditional probability:
$$\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}$$
Since $P(A\cap B)\neq 0$ for conditional probabilities to be defined, we can cancel it:
$$\frac{1}{P(B)}=\frac{1}{P(A)}$$
Thus,
$$P(A)=P(B)$$
Final Result
$$\boxed{P(A)=P(B)}$$
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Summary
Chapter 13 on Probability in NCERT Class 12 Mathematics Part II covers fundamental concepts of probability theory, including sample spaces, events, and probability calculations. It explores various probability scenarios, from simple experiments like coin tosses and die rolls to more complex situations involving multiple events and conditional probabilities. The chapter emphasizes the practical application of probability in real-world situations and provides a foundation for more advanced statistical concepts.
