Exercise 13.3 Class 12 Maths Probability

NCERT Solutions Class 12 Maths Chapter 13 Ex 13.3 | LearnCBSEHub

💡 The Formula

If $E_1, E_2, \dots, E_n$ are mutually exclusive and exhaustive events, and $A$ is any event, then the probability of $E_i$ given that $A$ has occurred is:

$$ P(E_i|A) = \frac{P(E_i) P(A|E_i)}{\sum_{j=1}^n P(E_j) P(A|E_j)} $$

Think of it as: (Path Probability of Specific Branch) / (Sum of Path Probabilities of All Branches).

Question 01

Urn: 5 Red, 5 Black. Draw ball, return it + 2 same color balls. Draw 2nd ball. Prob(2nd Red)?

Step 1: Define Events (1st Draw)

$E_1$: 1st ball is Red. $P(E_1) = 5/10 = 1/2$.
$E_2$: 1st ball is Black. $P(E_2) = 5/10 = 1/2$.
Let A: 2nd ball is Red.

Step 2: Conditional Probabilities (2nd Draw)

If $E_1$ (Red drawn): Urn has $5+2=7$ Red, 5 Black. Total 12.
$P(A|E_1) = 7/12$.
If $E_2$ (Black drawn): Urn has 5 Red, $5+2=7$ Black. Total 12.
$P(A|E_2) = 5/12$.

Step 3: Total Probability

$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$
$= \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2}$.

Answer: 1/2

Question 02

Bag I: 4R, 4B. Bag II: 2R, 6B. Bag selected at random, ball is Red. Find Prob(Bag I).

$E_1$: Bag I selected ($P(E_1)=1/2$). $E_2$: Bag II selected ($P(E_2)=1/2$).
A: Ball drawn is Red.
$P(A|E_1) = 4/8 = 1/2$.
$P(A|E_2) = 2/8 = 1/4$.
Bayes’ Theorem: $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
$= \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}} = \frac{1/4}{1/4 + 1/8} = \frac{1/4}{3/8} = \frac{2}{3}$.

Answer: 2/3

Question 03

Hostel: 60%, Day Scholar: 40%. Grade A: 30% (Hostel), 20% (Day). Selected student has Grade A. Prob(Hostelier)?

$E_1$: Hostelier, $P(E_1) = 0.6$. $E_2$: Day Scholar, $P(E_2) = 0.4$.
A: Attains Grade A.
$P(A|E_1) = 0.3$. $P(A|E_2) = 0.2$.
$P(E_1|A) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26} = \frac{9}{13}$.

Answer: 9/13

Question 04

Student answers question. Knows: 3/4, Guesses: 1/4. Correct guess prob: 1/4. Answer is correct. Prob(Knew answer)?

$E_1$: Knows answer, $P(E_1)=3/4$. $E_2$: Guesses, $P(E_2)=1/4$.
A: Answer is correct.
$P(A|E_1) = 1$ (If knows, definitely correct).
$P(A|E_2) = 1/4$ (Given).
$P(E_1|A) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{3/4}{3/4 + 1/16} = \frac{3/4}{13/16} = \frac{12}{13}$.

Answer: 12/13

Question 05

Disease Test. Present(0.1%), Test Positive(99%). Healthy(99.9%), False Positive(0.5%). Test Positive. Prob(Has Disease)?

$E_1$: Has Disease, $P(E_1) = 0.001$. $E_2$: Healthy, $P(E_2) = 0.999$.
A: Test Result Positive.
$P(A|E_1) = 99\% = 0.99$.
$P(A|E_2) = 0.5\% = 0.005$.
$P(E_1|A) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{99}{99 + 4995} = \frac{99}{5094} = \frac{22}{1132}$. Actually $99 / (99 + 999*5) = 99 / (99 + 4995) = 99/5094 = 11/566$? Wait.
Calculation: $\frac{0.00099}{0.00099 + 0.004995} = \frac{990}{990+4995} = \frac{990}{5985} = \frac{22}{133}$.

Answer: 22/133

Question 06

3 Coins: Two-headed, Biased (75% H), Unbiased. Coin chosen, shows Heads. Prob(Two-headed)?

$E_1$: Two-headed, $E_2$: Biased, $E_3$: Unbiased. $P(E_i) = 1/3$.
A: Shows Heads.
$P(A|E_1) = 1$.
$P(A|E_2) = 0.75 = 3/4$.
$P(A|E_3) = 0.5 = 1/2$.
$P(E_1|A) = \frac{\frac{1}{3}(1)}{\frac{1}{3}(1 + \frac{3}{4} + \frac{1}{2})} = \frac{1}{9/4} = \frac{4}{9}$.

Answer: 4/9

Question 07

Insured: 2000 Scooter, 4000 Car, 6000 Truck. Accident Probs: 0.01, 0.03, 0.15. Accident occurs. Prob(Scooter)?

Total drivers = 12000.
$P(E_1)$ (Scooter) $= 2000/12000 = 1/6$. $P(A|E_1) = 0.01$.
$P(E_2)$ (Car) $= 4000/12000 = 2/6$. $P(A|E_2) = 0.03$.
$P(E_3)$ (Truck) $= 6000/12000 = 3/6$. $P(A|E_3) = 0.15$.
$P(E_1|A) = \frac{\frac{1}{6}(0.01)}{\frac{1}{6}(0.01 + 2(0.03) + 3(0.15))} = \frac{0.01}{0.01+0.06+0.45} = \frac{1}{52}$.

Answer: 1/52

Question 08

Machine A (60%, 2% def), Machine B (40%, 1% def). Item defective. Prob(Machine B)?

$P(E_1)=0.6, P(A|E_1)=0.02$.
$P(E_2)=0.4, P(A|E_2)=0.01$.
$P(E_2|A) = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.012 + 0.004} = \frac{4}{16} = \frac{1}{4}$.

Answer: 1/4

Question 09

Group 1 wins (0.6, new prod 0.7). Group 2 wins (0.4, new prod 0.3). New product introduced. Prob(Group 2)?

$P(E_1)=0.6, P(A|E_1)=0.7$.
$P(E_2)=0.4, P(A|E_2)=0.3$.
$P(E_2|A) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{12}{54} = \frac{2}{9}$.

Answer: 2/9

Question 10

Die roll. 5,6 $\to$ coin 3 times. 1,2,3,4 $\to$ coin once. Exactly one Head obtained. Prob(1,2,3 or 4)?

$E_1$ (5,6): $P(E_1) = 2/6 = 1/3$.
$E_2$ (1,2,3,4): $P(E_2) = 4/6 = 2/3$.
A: Exactly one Head.
If $E_1$ (3 tosses): Outcomes {HTH, HHT, THH…}. $P(A|E_1) = 3/8$.
If $E_2$ (1 toss): Outcomes {H, T}. $P(A|E_2) = 1/2$.
$P(E_2|A) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{1/3}{1/8 + 1/3} = \frac{1/3}{11/24} = \frac{8}{11}$.

Answer: 8/11

Question 11

Operators A(50%, 1% def), B(30%, 5% def), C(20%, 7% def). Defective found. Prob(Produced by A)?

$P(E_1)=0.5, P(A|E_1)=0.01$.
$P(E_2)=0.3, P(A|E_2)=0.05$.
$P(E_3)=0.2, P(A|E_3)=0.07$.
$P(E_1|A) = \frac{0.5(0.01)}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34}$.

Answer: 5/34

Question 12

Card lost. Two drawn from remainder are Diamonds. Prob(Lost card is Diamond)?

Step 1: Events

$E_1$: Lost card is Diamond. $P(E_1) = 13/52 = 1/4$.
$E_2$: Lost card is Not Diamond. $P(E_2) = 39/52 = 3/4$.
A: Two cards drawn are Diamonds from remaining 51.

Step 2: Conditionals

If $E_1$: 12 Diamonds left. $P(A|E_1) = {}^{12}C_2 / {}^{51}C_2$.
If $E_2$: 13 Diamonds left. $P(A|E_2) = {}^{13}C_2 / {}^{51}C_2$.

Step 3: Bayes

$P(E_1|A) = \frac{\frac{1}{4} {}^{12}C_2}{\frac{1}{4} {}^{12}C_2 + \frac{3}{4} {}^{13}C_2}$.
${}^{12}C_2 = 66, {}^{13}C_2 = 78$.
$= \frac{1 \cdot 66}{1 \cdot 66 + 3 \cdot 78} = \frac{66}{66 + 234} = \frac{66}{300} = \frac{11}{50}$.

Answer: 11/50

Question 13

A speaks truth 4/5 times. Coin tossed. A reports Head. Prob(Actually Head)?

$E_1$: Head occurs (1/2). $E_2$: Tail occurs (1/2).
A: Reports Head.
$P(A|E_1)$: Head occurs, Reports Head (Truth) = 4/5.
$P(A|E_2)$: Tail occurs, Reports Head (Lie) = 1/5.
$P(E_1|A) = \frac{\frac{1}{2}(4/5)}{\frac{1}{2}(4/5) + \frac{1}{2}(1/5)} = \frac{4}{5}$.

Correct Option: (A) 4/5

Question 14

If $A \subset B$ and $P(B) \neq 0$, then which is correct?

$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Since $A \subset B$, $A \cap B = A$.
So $P(A|B) = \frac{P(A)}{P(B)}$.
Since $P(B) \le 1$, $\frac{P(A)}{P(B)} \ge P(A)$.
Thus, $P(A|B) \ge P(A)$.

Correct Option: (C)