Class 12 Ch 13: Probability - Exercise 13.2

Q1

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If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$, find $P(A \cap B)$ if A and B are independent events.

Since A and B are independent, $P(A \cap B) = P(A) \cdot P(B)$.

$P(A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$.

$\boxed{\frac{3}{25}}$

Q2

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Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Total cards = 52. Total black cards = 26.

$P(\text{First Black}) = \frac{26}{52} = \frac{1}{2}$.

Since drawn without replacement, remaining cards = 51, remaining black = 25.

$P(\text{Second Black}) = \frac{25}{51}$.

$P(\text{Both Black}) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$.

$\boxed{\frac{25}{102}}$

Q3

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A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Total = 15, Good = 12. We need to draw 3 good oranges consecutively without replacement.

$P(\text{All 3 Good}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$.

$= \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13} = \frac{4 \times 11 \times 2}{14 \times 13} = \frac{4 \times 11}{7 \times 13} = \frac{44}{91}$.

$\boxed{\frac{44}{91}}$

Q4

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A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.

Sample space size = $2 \times 6 = 12$.

$A = \{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6)\}$. $P(A) = \frac{6}{12} = \frac{1}{2}$.

$B = \{(H,3), (T,3)\}$. $P(B) = \frac{2}{12} = \frac{1}{6}$.

$A \cap B = \{(H,3)\}$. $P(A \cap B) = \frac{1}{12}$.

$P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

Since $P(A \cap B) = P(A)P(B)$, events are independent.

$\boxed{\text{Independent}}$

Q7

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Given that the events A and B are such that $P(A) = \frac{1}{2}, P(A \cup B) = \frac{3}{5}$ and $P(B) = p$. Find p if they are (i) mutually exclusive (ii) independent.

Formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. So $\frac{3}{5} = \frac{1}{2} + p - P(A \cap B)$.

(i) Mutually exclusive $\Rightarrow P(A \cap B) = 0$.
$\frac{3}{5} = \frac{1}{2} + p \Rightarrow p = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$.

(ii) Independent $\Rightarrow P(A \cap B) = P(A)P(B) = \frac{1}{2}p$.
$\frac{3}{5} = \frac{1}{2} + p - \frac{1}{2}p \Rightarrow \frac{3}{5} - \frac{1}{2} = \frac{1}{2}p \Rightarrow \frac{1}{10} = \frac{p}{2} \Rightarrow p = \frac{1}{5}$.

$\boxed{\text{(i) } \frac{1}{10}, \text{ (ii) } \frac{1}{5}}$

Q8

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Let A and B be independent events with $P(A) = 0.3$ and $P(B) = 0.4$. Find (i) $P(A \cap B)$ (ii) $P(A \cup B)$ (iii) $P(A|B)$ (iv) $P(B|A)$.

(i) $P(A \cap B) = P(A)P(B) = 0.3 \times 0.4 = 0.12$.

(ii) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.12 = 0.58$.

(iii) $P(A|B) = P(A) = 0.3$ (since independent).

(iv) $P(B|A) = P(B) = 0.4$ (since independent).

$\boxed{0.12, 0.58, 0.3, 0.4}$

Q11

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Given two independent events A and B such that $P(A) = 0.3, P(B) = 0.6$. Find (i) $P(A \text{ and } B)$ (ii) $P(A \text{ and not } B)$ (iii) $P(A \text{ or } B)$ (iv) $P(\text{neither A nor B})$.

(i) $P(A \cap B) = 0.3 \times 0.6 = 0.18$.

(ii) $P(A \cap B') = P(A) - P(A \cap B) = 0.3 - 0.18 = 0.12$.

(iii) $P(A \cup B) = 0.3 + 0.6 - 0.18 = 0.72$.

(iv) $P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.72 = 0.28$.

$\boxed{0.18, 0.12, 0.72, 0.28}$

Q12

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A die is tossed thrice. Find the probability of getting an odd number at least once.

$P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$. $P(\text{Even}) = \frac{1}{2}$.

$P(\text{At least one odd}) = 1 - P(\text{None odd}) = 1 - P(\text{All even})$.

$P(\text{All even}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

Required Probability = $1 - \frac{1}{8} = \frac{7}{8}$.

$\boxed{\frac{7}{8}}$

Q13

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Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.

Total = 18. $P(B) = \frac{10}{18} = \frac{5}{9}, P(R) = \frac{8}{18} = \frac{4}{9}$.

(i) $P(RR) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$.

(ii) $P(BR) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$.

(iii) $P(BR \cup RB) = P(BR) + P(RB) = \frac{20}{81} + (\frac{4}{9} \times \frac{5}{9}) = \frac{40}{81}$.

$\boxed{\frac{16}{81}, \frac{20}{81}, \frac{40}{81}}$

Q14

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Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}$. $P(A') = \frac{1}{2}, P(B') = \frac{2}{3}$.

(i) Problem solved = $1 - P(\text{None solves}) = 1 - P(A')P(B') = 1 - (\frac{1}{2} \times \frac{2}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$.

(ii) Exactly one = $P(A)P(B') + P(A')P(B) = (\frac{1}{2} \times \frac{2}{3}) + (\frac{1}{2} \times \frac{1}{3}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

$\boxed{\frac{2}{3}, \frac{1}{2}}$

Q17

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The probability of obtaining an even prime number on each die, when a pair of dice is rolled is: (A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Even prime number is 2.

We need 2 on first die AND 2 on second die.

Probability = $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.

$\boxed{\text{(D) } \frac{1}{36}}$

Q18

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Two events A and B will be independent, if:
(A) A and B are mutually exclusive
(B) $P(A'B') = [1 - P(A)][1 - P(B)]$
(C) $P(A) = P(B)$
(D) $P(A) + P(B) = 1$

If A and B are independent, then A' and B' are also independent.

So, $P(A' \cap B') = P(A')P(B') = [1 - P(A)][1 - P(B)]$.

$\boxed{\text{(B)}}$

Exercise 13.2 Practice