Anand Classes NCERT Solutions Probability Exercise 13.3 Class 12 Chapter-13 Math Notes PDF Free Download (Set-2) provides students with clear, step-by-step explanations to strengthen conceptual understanding and improve problem-solving accuracy. These solutions follow the latest NCERT guidelines and are ideal for Class 12 students preparing for board exams, competitive exams, or revision sessions. The notes are designed to make probability concepts easier, more intuitive, and exam-oriented for quick learning. Click the print button to download study material and notes.
Access Probability NCERT Solutions for Exercise 13.3 Class 12 Chapter 13
NCERT Question 8: A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Solution:
Let $A$ : Items produced by machine A
$B$ : Items produced by machine B
$D$ : Item is defective
We need to find the probability that the item was produced by machine $B$ if it is found to be defective, i.e. $P(B|D)$.
Using
$$P(B|D)=\frac{P(B)P(D|B)}{P(B)P(D|B)+P(A)P(D|A)}$$
Given
$P(A)=0.6$
$P(D|A)=0.02$
$P(B)=0.4$
$P(D|B)=0.01$
Substituting values,
$$P(B|D)=\frac{0.4\times 0.01}{0.6\times 0.02+0.4\times 0.01}$$
$$P(B|D)=\frac{0.01\times 0.4}{0.01(0.6 \times 2+0.4)}$$
$$P(B|D)=\frac{0.4}{1.2+0.4}$$
$$P(B|D)=\frac{0.4}{1.6}$$
$$P(B|D)=\frac{4}{16}=\frac{1}{4}$$
Final Result:
$$\boxed{\frac{1}{4}}$$
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NCERT Question.9: Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution:
Let $E$ : first group wins
$F$ : second group wins
$G$ : new product is introduced
We need to find the Probability that the new product introduced was by the second group i.e. $P(F|G)$.
We need $P(F|G)$.
$$P(F|G)=\frac{P(F)P(G|F)}{P(E)P(G|E)+P(F)P(G|F)}$$
Given
$P(E)=0.6$
$P(G|E)=0.7$
$P(F)=0.4$
$P(G|F)=0.3$
Substituting,
$$P(F|G)=\frac{0.4 \times 0.3}{0.6\times 0.7+0.4\times 0.3}$$
$$P(F|G)=\frac{0.12}{0.42+0.12}$$
$$P(F|G)=\frac{0.12}{0.54}=\frac{2}{9}$$
Final Result:
$$\boxed{\frac{2}{9}}$$
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NCERT Question.10 : Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution
Let
A : 1, 2, 3, 4 appear on the die
B : 5, 6 appear on the die
C : exactly one head is obtained
We need to find the Probability that if exactly one head is obtained on the toss of a coin, she threw 1, 2, 3 or 4 with the die i.e. $P(A \mid C)$
We need to find
$$P(A \mid C)$$
Using
$$P(A \mid C) = \frac{P(A)P(C \mid A)}{P(A)P(C \mid A) + P(B)P(C \mid B)}$$
Compute each probability:
Probability that 1, 2, 3 or 4 appear:
$$P(A) = \frac{4}{6} = \frac{2}{3}$$
If A occurs she tosses one coin so probability of exactly one head:
$$P(C \mid A) = \frac{1}{2}$$
Probability that 5 or 6 appear:
$$P(B) = \frac{2}{6} = \frac{1}{3}$$
If B occurs she tosses three coins so probability of exactly one head:
Number of ways to get exactly one head out of three tosses = 3
Total possible outcomes = 8
$$P(C \mid B) = \frac{3}{8}$$
Substitute values:
$$P(A \mid C) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{2}{3}\times \frac{1}{2} + \frac{1}{3}\times \frac{3}{8}}$$
Simplify numerator:
$$\frac{2}{3}\times \frac{1}{2} = \frac{1}{3}$$
Simplify denominator:
$$\frac{2}{3} \times\frac{1}{2} = \frac{1}{3}$$
$$\frac{1}{3} \times\frac{3}{8} = \frac{3}{24} = \frac{1}{8}$$
So denominator becomes
$$\frac{1}{3} + \frac{1}{8} = \frac{8}{24} + \frac{3}{24} = \frac{11}{24}$$
Thus
$$P(A \mid C) = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{1}{3} \times\frac{24}{11} = \frac{8}{11}$$
Final Result
$$\boxed{\frac{8}{11}}$$
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NCERT Question.11 : A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?
Solution
Let
A : item produced by operator A
B : item produced by operator B
C : item produced by operator C
D : item produced is defective
We need to find out the Probability that item is produced by operator A if it is defective i.e. $P(A \mid D)$.
We need to find
$$P(A \mid D)$$
Using
$$P(A \mid D) = \frac{P(A)P(D \mid A)}{P(A)P(D \mid A) + P(B)P(D \mid B) + P(C)P(D \mid C)}$$
Compute each probability:
Probability that item is produced by operator A:
$$P(A) = 0.5$$
Probability of defective item if produced by A:
$$P(D \mid A) = 0.01$$
Probability that item is produced by operator B:
$$P(B) = 0.3$$
Probability of defective item if produced by B:
$$P(D \mid B) = 0.05$$
Probability that item is produced by operator C:
$$P(C) = 0.2$$
Probability of defective item if produced by C:
$$P(D \mid C) = 0.07$$
Substitute values:
$$P(A \mid D) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07}$$
Simplify numerator:
$$0.5 \times 0.01 = 0.005$$
Simplify denominator:
$$0.5 \times 0.01 = 0.005$$
$$0.3 \times 0.05 = 0.015$$
$$0.2 \times 0.07 = 0.014$$
So denominator becomes
$$0.005 + 0.015 + 0.014 = 0.034$$
Thus
$$P(A \mid D) = \frac{0.005}{0.034} = \frac{5}{34}$$
Final Result
$$\boxed{\frac{5}{34}}$$
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NCERT Question.12 : A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Solution
Let
E1 : lost card is a diamond
E2 : lost card is not a diamond
A : both drawn cards are diamonds
We need to find out the probability that the lost card being a diamond if two cards drawn are found to be both diamond. i.e. $P(E1 \mid A)$
We need
$$P(E1 \mid A)$$
Using
$$P(E1 \mid A) = \frac{P(E1)P(A \mid E1)}{P(E1)P(A \mid E1) + P(E2)P(A \mid E2)}$$
Probability that the lost card is a diamond:
$$P(E1) = \frac{13}{52} = \frac14$$
If the lost card is a diamond, only 12 diamonds remain out of 51 cards.
Number of ways to choose 2 diamonds out of 12:
$$^{12}C_2 = \frac{12 \times 11}{2}$$
Number of ways to choose any 2 cards out of 51:
$$^{51}C_2 = \frac{51 \times 50}{2}$$
Thus,
$$P(A \mid E1) = \frac{^{12}C_2}{^{51}C_2} = \frac{12 \times 11}{51 \times 50}$$
If the lost card is not a diamond, all 13 diamonds remain.
Number of ways to choose 2 diamonds from 13:
$$^{13}C_2 = \frac{13 \times 12}{2}$$
Thus,
$$P(A \mid E2) = \frac{^{13}C_2}{^{51}C_2} = \frac{13 \times 12}{51 \times 50}$$
Probability that the lost card is not a diamond:
$$P(E2) = 1 – \frac14 = \frac34$$
Substituting in Bayes’ Formula
$$
P(E1 \mid A) =
\frac{\frac14 \times \frac{12 \times 11}{51 \times 50}}
{\frac14 \times \frac{12 \times 11}{51 \times 50} + \frac34 \times \frac{13 \times 12}{51 \times 50}}
$$
Factor out the common term $(\frac{12}{51 \times 50})$:
$$
P(E1 \mid A) =
\frac{\frac14 \times 11}
{\frac14 \times 11 + \frac34 \times 13}
$$
Simplify:
$$
P(E1 \mid A) = \frac{11}{11 + 39} = \frac{11}{50}
$$
Final Result
$$\boxed{\frac{11}{50}}$$
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NCERT Question.13 : Probability that A speaks truth is $ \frac45 $. A coin is tossed. A reports that a head appears. The probability that actually there was head is
(A) $ \frac45 $
(B) $ \frac12 $
(C) $ \frac15 $
(D) $ \frac25 $
Solution
Let
$E$ : A speaks truth
$F$ : A lies
$H$ : head actually appears
We need to find the Probability that head actually appears, if A reports that a head appears i.e. $P(H \mid \text{A reports head})$.
We need
$$P(H \mid \text{A reports head})$$
Since A reports head, this report can come from:
- A speaks truth and head actually appears
- A lies and tail appears
Thus,
$$P(H \mid \text{report head}) = \frac{P(E)P(H \mid E)}{P(E)P(H \mid E) + P(F)P(H \mid F)}$$
Given:
$$P(E) = \frac45$$
$$P(F) = 1 – \frac45 = \frac15$$
If A speaks truth, he reports the actual face, so
$$P(H \mid E) = \frac12$$
If A lies, he reports the opposite face; probability of actual head when lying:
Tail occurs with probability $\frac12$, and A reports head instead.
Thus,
$$P(H \mid F) = \frac12$$
Substituting,
$$
P(H \mid \text{report head}) =
\frac{\frac45 \times\frac12}
{\frac45 \times\frac12 + \frac15 \times\frac12}
$$
Factor out $\frac12$:
$$
P(H \mid \text{report head}) = \frac{\frac45}{\frac45 + \frac15}
$$
$$
P(H \mid \text{report head}) = \frac{\frac45}{1}
$$
$$
P(H \mid \text{report head}) = \frac45
$$
Final Result
$$\boxed{\frac45}$$
Correct option: (A)
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NCERT Question.14 : If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct?
(A) $P(A \mid B) = \dfrac{P(B)}{P(A)}$
(B) $P(A \mid B) < P(A)$
(C) $P(A \mid B) \ge P(A)$
(D) None of these
Solution
Given that
$$A \subset B$$
So,
$$A \cap B = A$$
$$P(A \cap B) = P(A)$$
Using conditional probability:
$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}
$$
Since
$$0 < P(B) \le 1$$
dividing by $P(B)$ (which is at most 1) increases or keeps the same value:
$$
P(A \mid B) = \frac{P(A)}{P(B)}
$$
$$
\Rightarrow \frac{P(A)}{P(B)} \ge P(A)
$$
because $P(B) \le 1$.
Thus,
$$
P(A \mid B) \ge P(A)
$$
Final Result
$$\boxed{P(A \mid B) \ge P(A)}$$
Correct option: (C)
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Summary
Exercise 13.3 focuses on the concept of random variables and probability distributions. It introduces discrete random variables, their probability mass functions (PMF), and cumulative distribution functions (CDF). Students learn to calculate expected values, variances, and standard deviations of discrete random variables. The exercise also covers properties of probability distributions and how to solve problems involving real-life applications of random variables.
