Question

If the p.m.f. of a random variable \( X \) is \[ \begin{array}{|c|c|c|c|c|} X & 1 & 2 & 3 & 4 & 5
P(X = x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \end{array} \] then \( k = \)

• A. \( \frac{15}{31} \)
• B. \( \frac{1}{12} \)
• C. \( \frac{11}{12} \)
• D. \( \frac{12}{31} \)

Hint:

For probability mass functions, ensure that the sum of probabilities equals 1 and solve for the constant \( k \).

Answer 1

The Correct Option is D

Step 1: Sum of probabilities must equal 1.
The sum of the probabilities for all possible values of \( X \) must be equal to 1: \[ k + \frac{k}{3} + \frac{k}{4} + \frac{k}{2} = 1 \]
Step 2: Solve for \( k \).
To simplify the equation, find the common denominator: \[ k \left( 1 + \frac{1}{3} + \frac{1}{4} + \frac{1}{2} \right) = 1 \] \[ k \left( \frac{12}{12} + \frac{4}{12} + \frac{3}{12} + \frac{6}{12} \right) = 1 \] \[ k \times \frac{25}{12} = 1 \quad \Rightarrow \quad k = \frac{12}{31} \]
Step 3: Conclusion.
Thus, the value of \( k \) is \( \frac{12}{31} \), corresponding to option (D).