Maths PROBABILITY EXERCISE 13.2 Solution
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Maths PROBABILITY EXERCISE 13.2 Solution

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY EXERCISE 13.2 Questions 1 to 18

Probability Chapter 13 Exercises 13.2 Solutions

  1. To find the probability of the intersection of independent events A and B:

    Probability of A ∩ B = Probability of A × Probability of B

    So, P(A ∩ B) = P(A) × P(B) = 3/5 × 1/5 = 3/25

  2. To calculate the probability of drawing two black cards from a deck of 52 cards:

    Probability of both black = Probability of black on first draw × Probability of black on second draw

    Hence, P(both black) = P(black on first draw) × P(black on second draw) = 26/52 × 25/51 = 25/102

  3. To determine the probability of approving a box with three good oranges out of 15:

    Probability of approval = Number of favorable outcomes / Total number of outcomes

    So, P(approved) = 220/455

  4. Events A and B are independent if P(A ∩ B) = P(A) × P(B).

    Given P(A ∩ B) = 1/12 and P(A) × P(B) = 1/2 × 1/6 = 1/12, A and B are independent.

  5. For events A ('the number is even') and B ('the number is red'), where P(A) = P(B) = 1/2:

    Probability of their intersection is 0, while P(A) × P(B) = 1/2 × 1/2 = 1/4.

    Thus, A and B are not independent.

  6. To determine if events E and F are independent:

    Compare P(E ∩ F) to P(E) × P(F):

    P(E ∩ F) = 1/5

    P(E) × P(F) = (3/5) × (3/10) = 9/50

    Since P(E ∩ F) ≠ P(E) × P(F), events E and F are not independent.

  7. Given events A and B such that P(A) = 1/2, P(A∪B) = 3/5, and P(B) = p, find p if they are:

    1. Mutually exclusive:

      If A and B are mutually exclusive, then P(A∪B) = P(A) + P(B).

      Given P(A∪B) = 3/5 and P(A) = 1/2, solving for p:

      p = 3/5 − 1/2 = 1/10

    2. Independent:

      If A and B are independent, then P(A∩B) = P(A) × P(B).

      Given P(A) = 1/2 and P(B) = p, and P(A∩B) = 0 (since they are mutually exclusive), solving for p:

      p = 0

  8. Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. We are asked to find:

    1. P(A∩B): P(A∩B) = P(A) × P(B) = 0.3 × 0.4 = 0.12
    2. P(A∪B): P(A∪B) = P(A) + P(B) − P(A∩B) = 0.3 + 0.4 − 0.12 = 0.58
    3. P(A|B): P(A|B) = P(A) since A and B are independent, so P(A|B) = 0.3
    4. P(B|A): P(B|A) = P(B) since A and B are independent, so P(B|A) = 0.4

  9. If A and B are two events such that P(A) = 1/4, P(B) = 1/2, and P(A∩B) = 1/8, we are asked to find P(not A and not B).

    P(not A and not B) = P(not A) × P(not B) = 3/4 × 1/2 = 3/8

  10. Events A and B are such that P(A) = 1/2, P(B) = 7/12, and P(not A or not B) = 1/4. We need to determine whether A and B are independent.

    Since P(not A or not B) represents the probability of the union of the complements of A and B, we don't have enough information to directly determine the independence of A and B from the given data.

  11. Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6, we are asked to find:

    1. P(A∩B): P(A∩B) = P(A) × P(B) = 0.3 × 0.6 = 0.18
    2. P(A∩notB): P(A∩notB) = P(A) × P(notB) = 0.3 × (1−0.6) = 0.3 × 0.4 = 0.12
    3. P(A∪B): P(A∪B) = P(A) + P(B) − P(A∩B) = 0.3 + 0.6 − 0.18 = 0.72
    4. P(neither A nor B): P(neither A nor B) = P(not A) × P(not B) = (1−0.3) × (1−0.6) = 0.7 × 0.4 = 0.28

  12. A die is tossed thrice. To find the probability of getting an odd number at least once, we can use the complementary probability approach. The probability of not getting an odd number in a single toss is 1/2, so the probability of not getting an odd number in three tosses is (1/2)3 = 1/8. Therefore, the probability of getting at least one odd number in three tosses is 1 − 1/8 = 7/8.

  13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. We are asked to find the probability that:

    1. Both balls are red: The probability of drawing two red balls is (8/18)2 = 16/81.
    2. The first ball is black and the second is red: The probability of this event is 5/9 × 4/9 = 20/81.
    3. One of them is black and the other is red: The total probability is 40/162 + 40/162 = 162/80.

  14. Probability of solving a specific problem independently by A and B are 1/2 and 1/3 respectively. We need to find the probability that:

    1. The problem is solved: P(solved) = 1 − P(not solved) = 1 − P(not A) × P(not B). Since A and B are independent, we have P(not A) = 1 − P(A) = 1−1/2 and P(not B) = 1 − P(B) = 2/3. Therefore, P(solved) = 1 − (1/2 × 2/3) = 1 − 1/3 = 2/3.
    2. Exactly one of them solves the problem: P(exactly one) = P(A) × P(not B) + P(not A) × P(B). Since A and B are independent, we have P(not B) = 2/3 and P(not A) = 1/2. Therefore, P(exactly one) = (1/2 × 2/3) + (1/2 × 1/3) = 1/3 + 1/6 = 1/2.

  15. One card is drawn at random from a well-shuffled deck of 52 cards. In each of the following cases, we determine whether events E and F are independent:

    1. E : ‘the card drawn is a spade’ F : ‘the card drawn is an ace’ Since there are 13 spades in the deck and 4 aces, but only 1 ace of spades, we have P(E ∩ F) = 1/52, P(E) = 13/52, and P(F) = 4/52. Since P(E ∩ F) ≠ P(E) × P(F), events E and F are not independent.
    2. E : ‘the card drawn is black’ F : ‘the card drawn is a king’ There are 26 black cards in the deck, and 2 black kings. So, P(E ∩ F) = 2/52, P(E) = 26/52, and P(F) = 4/52. Since P(E ∩ F) ≠ P(E) × P(F), events E and F are not independent.
    3. E : ‘the card drawn is a king or queen’ F : ‘the card drawn is a queen or jack’ There are 8 cards that are either a king or queen, and 8 cards that are either a queen or jack. Since there are 4 cards that are both a queen and a king, we have P(E ∩ F) = 4/52, P(E) = 8/52, and P(F) = 8/52. Since P(E ∩ F) ≠ P(E) × P(F), events E and F are not independent.

  16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper, and 20% read both Hindi and English newspapers. A student is selected at random.

    1. To find the probability that she reads neither Hindi nor English newspapers, we subtract the probability of reading either Hindi or English newspapers from 1: P(neither) = 1 − P(Hindi ∪ English) = 1 − (P(Hindi) + P(English) − P(Hindi ∩ English)) = 1 − (0.60 + 0.40 − 0.20) = 1 − 0.80 = 0.20.
    2. If she reads Hindi newspaper, the probability that she reads English newspaper is given by: P(English | Hindi) = P(English ∩ Hindi) / P(Hindi) = 0.20 / 0.60 = 1/3.
    3. If she reads English newspaper, the probability that she reads Hindi newspaper is given by: P(Hindi | English) = P(Hindi ∩ English) / P(English) = 0.20 / 0.40 = 1/2.
  17. The probability of obtaining an even prime number on each die when a pair of dice is rolled is:
    • (C) 1/12
  18. Two events A and B will be independent if:
    • (B) P(A′) × P(B′) = [1 − P(A)] × [1 − P(B)]