Question

Given below is the probability distribution of a discrete random variable \( X \):
\[ \begin{array}{c|cccccc} X=x & 1 & 2 & 3 & 4 & 5 & 6
\hline P(X=x) & k & 0 & 2k & 5k & k & 3k \end{array} \] Then find \( P(X \ge 4) \).

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{3} \)
• C. \( \dfrac{1}{2} \)
• D. \( \dfrac{3}{4} \)

Hint:

Always verify that the sum of probabilities in a discrete distribution is equal to 1 before calculating required probabilities.

Answer 1

The Correct Option is D

Step 1: Use the total probability condition.
\[ k + 0 + 2k + 5k + k + 3k = 1 \] \[ 12k = 1 \Rightarrow k = \frac{1}{12} \]

Step 2: Find \( P(X \ge 4) \).
\[ P(X \ge 4) = P(4) + P(5) + P(6) \] \[ = 5k + k + 3k = 9k \]

Step 3: Substitute the value of \( k \).
\[ P(X \ge 4) = 9 \times \frac{1}{12} = \frac{3}{4} \]

Step 4: Conclusion.
\[ \boxed{\frac{3}{4}} \]