Exercise 13.1 Class 12 Maths Probability

Questions 01 — 05

Basic Application of Formulas.

1. Given $P(E)=0.6, P(F)=0.3, P(E \cap F)=0.2$.

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}$
$P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$

2. Compute $P(A|B)$ if $P(B)=0.5, P(A \cap B)=0.32$.

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{0.64}{1} = 0.64$

3. Given $P(A)=0.8, P(B)=0.5, P(B|A)=0.4$.

(i) $P(A \cap B) = P(B|A) \cdot P(A) = 0.4 \times 0.8 = 0.32$
(ii) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$
(iii) $P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.8 + 0.5 – 0.32 = 0.98$

4. Evaluate $P(A \cup B)$ if $2P(A)=P(B)=\frac{5}{13}$ and $P(A|B)=\frac{2}{5}$.

$P(B) = \frac{5}{13}, \quad P(A) = \frac{5}{26}$
$P(A \cap B) = P(A|B) \cdot P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$
$P(A \cup B) = \frac{5}{26} + \frac{5}{13} – \frac{2}{13} = \frac{5 + 10 – 4}{26} = \frac{11}{26}$

5. Given $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}, P(A \cup B)=\frac{7}{11}$.

(i) $P(A \cap B) = P(A) + P(B) – P(A \cup B) = \frac{6+5-7}{11} = \frac{4}{11}$
(ii) $P(A|B) = \frac{4/11}{5/11} = \frac{4}{5}$
(iii) $P(B|A) = \frac{4/11}{6/11} = \frac{4}{6} = \frac{2}{3}$

Questions 06 — 09

Find $P(E|F)$ from Sample Spaces.

6. Coin tossed three times. $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$. Total = 8.

(i) E: Head on 3rd ($\{HHH, HTH, THH, TTH\}$), F: Heads on first two ($\{HHH, HHT\}$).
$E \cap F = \{HHH\}$. $P(E|F) = \frac{1/8}{2/8} = \frac{1}{2}$.

(ii) E: At least 2 heads (4 outcomes), F: At most 2 heads (7 outcomes, all except HHH).
$E \cap F = \{HHT, HTH, THH\}$. $P(E|F) = \frac{3/8}{7/8} = \frac{3}{7}$.

(iii) E: At most 2 tails (7 outcomes), F: At least 1 tail (7 outcomes).
$E \cap F$: Elements in both = $\{HHT, HTH, HTT, THH, THT, TTH\}$. (6 outcomes).
$P(E|F) = \frac{6/8}{7/8} = \frac{6}{7}$.

7. Two coins tossed once. $S = \{HH, HT, TH, TT\}$.

(i) E: Tail on one ($\{HT, TH\}$), F: One shows head ($\{HT, TH\}$).
$E = F$, so $P(E|F) = 1$.

(ii) E: No tail ($\{HH\}$), F: No head ($\{TT\}$).
$E \cap F = \phi$. $P(E|F) = 0$.

8. Die thrown 3 times. Total 216 outcomes.

F: 6 on first, 5 on second = $\{(6,5,1), (6,5,2), \dots, (6,5,6)\}$ (6 outcomes).
E: 4 on third.
$E \cap F = \{(6,5,4)\}$ (1 outcome).
$P(E|F) = \frac{1/216}{6/216} = \frac{1}{6}$.

9. Mother, Father, Son lineup. Total $3! = 6$.

$S = \{MFS, MSF, FMS, FSM, SMF, SFM\}$
E: Son on one end = $\{MFS, FMS, SMF, SFM\}$ (4 outcomes).
F: Father in middle = $\{MFS, SFM\}$ (2 outcomes).
$E \cap F = \{MFS, SFM\}$ (2 outcomes).
$P(E|F) = \frac{2/6}{2/6} = 1$.

Question 10

Two dice (Black & Red).

(a) Sum > 9, given Black = 5.

Let Black be 1st die. F (Black=5): $\{(5,1)…(5,6)\}$ (6 outcomes).
E (Sum > 9): $\{(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)\}$.
$E \cap F = \{(5,5), (5,6)\}$ (2 outcomes).
Probability = $\frac{2}{6} = \frac{1}{3}$.

(b) Sum = 8, given Red < 4.

Let Red be 2nd die. F (Red < 4): $\{(1,1), (1,2), (1,3), (2,1)...(6,3)\}$ (18 outcomes).
E (Sum = 8): $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$.
$E \cap F = \{(5,3), (6,2)\}$ (2 outcomes).
Probability = $\frac{2}{18} = \frac{1}{9}$.

Question 11

Fair die rolled. $E=\{1,3,5\}, F=\{2,3\}, G=\{2,3,4,5\}$.

(i) $P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{P(\{3\})}{P(\{2,3\})} = \frac{1/6}{2/6} = \frac{1}{2}$.
$P(F|E) = \frac{1/6}{3/6} = \frac{1}{3}$.

(ii) $P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{P(\{3,5\})}{P(\{2,3,4,5\})} = \frac{2/6}{4/6} = \frac{1}{2}$.
$P(G|E) = \frac{2/6}{3/6} = \frac{2}{3}$.

(iii) $E \cup F = \{1,2,3,5\}$. $(E \cup F) \cap G = \{2,3,5\}$.
$P((E \cup F)|G) = \frac{3/6}{4/6} = \frac{3}{4}$.
$E \cap F = \{3\}$. $(E \cap F) \cap G = \{3\}$.
$P((E \cap F)|G) = \frac{1/6}{4/6} = \frac{1}{4}$.

Question 12

Family with two children. Sample Space $S = \{BB, BG, GB, GG\}$.

Let Event A: Both are girls ($GG$).

(i) Youngest is a girl (Event B)

$B = \{GG, BG\}$. $A \cap B = \{GG\}$.
$P(A|B) = \frac{1}{2}$.

(ii) At least one is a girl (Event C)

$C = \{GG, GB, BG\}$. $A \cap C = \{GG\}$.
$P(A|C) = \frac{1}{3}$.

Question 13

Question Bank Probability.

Type	True/False	Multiple Choice	Total
Easy	300	500	800
Difficult	200	400	600
Total	500	900	1400

Let E: Easy Question, M: Multiple Choice Question.
Find $P(E|M)$.
$P(M) = \frac{900}{1400}$. $P(E \cap M) = \frac{500}{1400}$ (Easy AND MCQ).
$P(E|M) = \frac{500}{900} = \frac{5}{9}$.

Question 14

Two dice, numbers are different. Find prob sum is 4.

F (Different Numbers): Total $36 – 6$ (doubles) $= 30$ outcomes.
E (Sum is 4): $\{(1,3), (2,2), (3,1)\}$.
$E \cap F$: exclude $(2,2) \implies \{(1,3), (3,1)\}$ (2 outcomes).
$P(E|F) = \frac{2/36}{30/36} = \frac{2}{30} = \frac{1}{15}$.

Question 15

Die Experiment. Multiple of 3 $\to$ throw again. Else $\to$ toss coin.

Sample Space:
Branch 1 (3, 6): $\{(3,1)..(3,6), (6,1)..(6,6)\}$ (12 outcomes, prob $\frac{1}{6}\times\frac{1}{6}$ each).
Branch 2 (1,2,4,5): $\{(1,H), (1,T), (2,H), (2,T)…\}$ (8 outcomes, prob $\frac{1}{6}\times\frac{1}{2}$ each).

Event B (At least one die shows 3):
Includes $\{(3,1)…(3,6)\}$ and $\{(6,3)\}$. Total 7 outcomes.
$P(B) = \frac{6}{36} + \frac{1}{36} = \frac{7}{36}$. (Using weight of dice branch).

Event A (Coin shows Tail): $\{(1,T), (2,T), (4,T), (5,T)\}$.
$P(A) = 4 \times \frac{1}{12}$.

Intersection $A \cap B = \phi$ (Impossible to have coin tail AND die 3).
$P(A|B) = \frac{0}{P(B)} = 0$.

Questions 16 — 17

Multiple Choice Questions.

16. If $P(A) = 1/2, P(B) = 0$, then $P(A|B)$ is?

$P(A|B) = \frac{P(A \cap B)}{P(B)}$. Since denominator is 0, it is not defined.

Correct Option: (C) not defined

17. If $P(A|B) = P(B|A)$, then?

$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$
Assuming $P(A \cap B) \neq 0$, we get $P(A) = P(B)$.

Correct Option: (D) $P(A) = P(B)$

Conditional Probability

💡 Formula: Conditional Probability