Question

Consider the probability distribution:

Then the value of $P(X > 2)$ is ______.

• A. 7/12
• B. 1/36
• C. 1/2
• D. 23/36

Hint:

To save time on calculation, you can also compute $P(X > 2)$ using the complement rule: $P(X > 2) = 1 - [P(X=1) + P(X=2)] = 1 - (K + 2K) = 1 - 3K = 1 - 3(1/6) = 1 - 1/2 = 1/2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The sum of all probabilities in a distribution must equal 1.

Step 2: Formula Application:
$K + 2K + K^2 + 2K + 5K^2 = 1 \implies 6K^2 + 5K - 1 = 0$.

Step 3: Explanation:
Factorizing: $(6K - 1)(K + 1) = 0$. Since $P(x) \geq 0$, $K = 1/6$. $P(X > 2) = P(3) + P(4) + P(5) = K^2 + 2K + 5K^2 = 6K^2 + 2K$. Substituting $K = 1/6$: $6(1/36) + 2(1/6) = 1/6 + 2/6 = 3/6 = 1/2$. Wait, re-checking $P(X>2)$: $P(3)+P(4)+P(5) = (1/6)^2 + 2(1/6) + 5(1/6)^2 = 1/36 + 12/36 + 5/36 = 18/36 = 1/2$. (Note: If the distribution includes $X=5$ as $10K^2$, the result would be 23/36).

Step 4: Final Answer:
The value is 1/2 (based on the provided table) or 23/36 (based on common variations of this exam question).