Question

A random variable X has the following probability distribution :

X:	1	2	3	4	5
P(X):	\(K^2\)	\(2K\)	\(K\)	\(2K\)	\(5K^2\)

Then \(P(X \gt 2)\) is equal to :

• A. \(\frac{1}{6}\)
• B. \(\frac{7}{12}\)
• C. \(\frac{1}{36}\)
• D. \(\frac{23}{36}\)

Hint:

Use that the probabilities sum to 1 to find K (K must be positive), then add P(3), P(4), P(5).

Answer 1

The Correct Option is D

Sum of probabilities = 1: \(K^2 + 2K + K + 2K + 5K^2 = 6K^2 + 5K = 1\).
\(6K^2 + 5K - 1 = 0\), \((6K - 1)(K + 1) = 0\), so \(K = \frac{1}{6}\) (K = -1 rejected as probability cannot be negative).
\(P(X \gt 2) = P(3) + P(4) + P(5) = K + 2K + 5K^2 = 3K + 5K^2\).
\(= \frac{3}{6} + \frac{5}{36} = \frac{18}{36} + \frac{5}{36} = \frac{23}{36}\).
Answer: (4) \(\frac{23}{36}\)