Two different numbers are chosen at random from the numbers 1, 2, 3, …, 9. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Probability that both numbers are odd, given that their sum is even | (I) | \(\displaystyle\frac{13}{18}\) |
| (B) | Probability that the sum of the numbers is even | (II) | \(\displaystyle\frac{5}{8}\) |
| (C) | Probability that both numbers are even, given that their sum is even | (III) | \(\displaystyle\frac{4}{9}\) |
| (D) | Probability that the product of the numbers is even | (IV) | \(\displaystyle\frac{3}{8}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
The numbers 1 to 9 have 5 odd and 4 even numbers. All choices of two numbers are equally likely:
\({}^{9}C_{2} = 36\)
Both odd: \({}^{5}C_{2} = 10\)
Both even: \({}^{4}C_{2} = 6\)
The sum is even exactly when both numbers are odd or both are even, so 16 choices have an even sum. Given an even sum, the reduced sample space has these 16 choices.
(A) Probability that both numbers are odd, given that their sum is even
This matches (II) \(\frac{5}{8}\).
\( \displaystyle P = \frac{10}{16} = \frac{5}{8} \)
(B) Probability that the sum of the numbers is even
This matches (III) \(\frac{4}{9}\).
\( \displaystyle P = \frac{16}{36} = \frac{4}{9} \)
(C) Probability that both numbers are even, given that their sum is even
This matches (IV) \(\frac{3}{8}\).
\( \displaystyle P = \frac{6}{16} = \frac{3}{8} \)
(D) Probability that the product of the numbers is even
This matches (I) \(\frac{13}{18}\).
The product is odd only when both numbers are odd, so by the complement:
\( \displaystyle P = 1-\frac{10}{36} = \frac{26}{36} = \frac{13}{18} \)