Match List-I with List-II
| List-I | List-II | ||
| (A) | Probability that a card drawn at random from a well-shuffled pack of 52 cards is a king or a spade | (I) | \(\displaystyle\frac{3}{8}\) |
| (B) | Two dice are thrown. Probability that the sum of the numbers is 3, given that the sum is less than 6 | (II) | \(\displaystyle\frac{1}{10}\) |
| (C) | In a class, 40% of the students are girls and 25% of the girls play chess. Probability that a student chosen at random is a girl who plays chess | (III) | \(\displaystyle\frac{4}{13}\) |
| (D) | \(P(A'|B')\), where P(A) = 0.6, P(B) = 0.2 and P(A|B) = 0.5 | (IV) | \(\displaystyle\frac{1}{5}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) Probability that a card drawn at random from a well-shuffled pack of 52 cards is a king or a spade
This matches (III) \(\frac{4}{13}\).
There are 13 spades and 4 kings, and the king of spades is in both groups, so by the addition theorem it is counted once:
\( \displaystyle P = \frac{13+4-1}{52} = \frac{16}{52} = \frac{4}{13} \)
(B) Two dice are thrown. Probability that the sum of the numbers is 3, given that the sum is less than 6
This matches (IV) \(\frac{1}{5}\).
The given condition shrinks the sample space. Outcomes with sum less than 6:
sum 2: (1, 1)
sum 3: (1, 2), (2, 1)
sum 4: (1, 3), (2, 2), (3, 1)
sum 5: (1, 4), (2, 3), (3, 2), (4, 1)
That is 10 outcomes, of which 2 have sum 3:
\( \displaystyle P = \frac{2}{10} = \frac{1}{5} \)
(C) In a class, 40% of the students are girls and 25% of the girls play chess. Probability that a student chosen at random is a girl who plays chess
This matches (II) \(\frac{1}{10}\).
By the multiplication theorem, P(girl and chess) = P(girl) × P(chess | girl):
\( \displaystyle P = 0.4\times 0.25 = 0.1 = \frac{1}{10} \)
(D) \(P(A'|B')\), where P(A) = 0.6, P(B) = 0.2 and P(A|B) = 0.5
This matches (I) \(\frac{3}{8}\).
\(P(A\cap B) = 0.5\times 0.2 = 0.1\)
\(P(A\cup B) = 0.6+0.2-0.1 = 0.7\)
Since \(A'\cap B'\) is the complement of \(A\cup B\):
\(P(A'\cap B') = 1-0.7 = 0.3\)
\(P(B') = 1-0.2 = 0.8\)
\( \displaystyle P(A'|B') = \frac{P(A'\cap B')}{P(B')} = \frac{0.3}{0.8} = \frac{3}{8} \)