The probability distribution of a random variable \(X\) is:
\(P(X = x)=\begin{cases}kx^2, & \text{for } x=1,2,3\\ 2kx, & \text{for } x=4,5,6\\ 0, & \text{otherwise}\end{cases}\), where \(k\) is a constant.
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(k\) | (I) | 7/22 |
| (B) | \(P(X \ge 4)\) | (II) | 1/44 |
| (C) | \(P(X \lt 4)\) | (III) | 95/22 |
| (D) | \(E[X]\) | (IV) | 15/22 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
The probabilities are \(k, 4k, 9k\) for \(x = 1, 2, 3\) and \(8k, 10k, 12k\) for \(x = 4, 5, 6\).
(A) \(k\)
This matches (II) 1/44. The sum of all probabilities is 1:
\(k(1 + 4 + 9) + 2k(4 + 5 + 6) = 1\)
\(14k + 30k = 1\)
\( \displaystyle 44k = 1,\quad k = \frac{1}{44} \)
(B) \(P(X \ge 4)\)
This matches (IV) 15/22.
\( \displaystyle P(X \ge 4) = 8k + 10k + 12k = 30k = \frac{30}{44} = \frac{15}{22} \)
(C) \(P(X \lt 4)\)
This matches (I) 7/22.
\( \displaystyle P(X \lt 4) = k + 4k + 9k = 14k = \frac{14}{44} = \frac{7}{22} \)
(D) \(E[X]\)
This matches (III) 95/22. The expected value is \(E[X] = \sum x\,P(X = x)\):
\(E[X] = 1(k) + 2(4k) + 3(9k) + 4(8k) + 5(10k) + 6(12k)\)
\(= k + 8k + 27k + 32k + 50k + 72k = 190k\)
\( \displaystyle E[X] = \frac{190}{44} = \frac{95}{22} \)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).