Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | A coin is tossed twice. Probability of two heads, given that the first toss is a head | (I) | \(\displaystyle\frac{1}{4}\) |
| (B) | A die is thrown. Probability that the number is a prime, given that it is odd | (II) | \(\displaystyle\frac{1}{5}\) |
| (C) | P(A|B), where \(P(A\cap B) = 0.12\) and P(B) = 0.48 | (III) | \(\displaystyle\frac{1}{2}\) |
| (D) | P(A|B), where A and B are independent, \(P(A) = \displaystyle\frac{1}{5}\) and \(P(B) = \displaystyle\frac{3}{4}\) | (IV) | \(\displaystyle\frac{2}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) A coin is tossed twice. Probability of two heads, given that the first toss is a head
This matches (III) \(\frac{1}{2}\).
The condition leaves the reduced sample space {HH, HT}, and only HH has two heads:
\( \displaystyle P = \frac{1}{2} \)
(B) A die is thrown. Probability that the number is a prime, given that it is odd
This matches (IV) \(\frac{2}{3}\).
The odd numbers are 1, 3, 5. Of these, 3 and 5 are prime (1 is not a prime):
\( \displaystyle P = \frac{2}{3} \)
(C) P(A|B), where \(P(A\cap B) = 0.12\) and P(B) = 0.48
This matches (I) \(\frac{1}{4}\).
\( \displaystyle P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{0.12}{0.48} = \frac{1}{4} \)
(D) P(A|B), where A and B are independent, \(P(A) = \frac{1}{5}\) and \(P(B) = \frac{3}{4}\)
This matches (II) \(\frac{1}{5}\).
For independent events, \(P(A\cap B) = P(A)P(B)\), so the condition B does not change the chance of A:
\( \displaystyle P(A|B) = \frac{\frac{1}{5}\times\frac{3}{4}}{\frac{3}{4}} = \frac{1}{5} \)