If a random variable \(X\) has the following probability distribution:
| \(X\) | 0 | 1 | 2 | 3 | 4 |
| \(P(X)\) | \(k\) | \(2k\) | \(3k\) | \(k^2\) | \(6k^2\) |
, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(k\) | (I) | 3/7 |
| (B) | \(P(X \lt 2)\) | (II) | 6/49 |
| (C) | \(P(X \gt 3)\) | (III) | 1/7 |
| (D) | \(P(2 \le X \le 3)\) | (IV) | 22/49 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
The sum of all probabilities is 1:
\(k + 2k + 3k + k^2 + 6k^2 = 1\)
\(7k^2 + 6k - 1 = 0\)
\((7k - 1)(k + 1) = 0\)
A probability cannot be negative, so \(k = -1\) is rejected and \(k = \frac{1}{7}\).
(A) \(k\)
This matches (III) 1/7, since \(k = \frac{1}{7}\).
(B) \(P(X \lt 2)\)
This matches (I) 3/7.
\( \displaystyle P(X = 0) + P(X = 1) = k + 2k = 3k = \frac{3}{7} \)
(C) \(P(X \gt 3)\)
This matches (II) 6/49.
\( \displaystyle P(X = 4) = 6k^2 = \frac{6}{49} \)
(D) \(P(2 \le X \le 3)\)
This matches (IV) 22/49.
\( \displaystyle P(X = 2) + P(X = 3) = 3k + k^2 = \frac{3}{7} + \frac{1}{49} \)
\( \displaystyle = \frac{21 + 1}{49} = \frac{22}{49} \)
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).