Question

If a random variable \( x \) has the probability distribution 

then \( P(3<x \leq 6) \) is equal to

• A. 0.22
• B. 0.33
• C. 0.34
• D. 0.64

Hint:

Always use the normalization condition \( \sum P(x) = 1 \) first to find unknown constants in a probability distribution before calculating specific probabilities.

Answer 1

The Correct Option is D

To find \( P(3 < x \leq 6) \), we need to find the sum of probabilities for \( x = 4, 5, \) and \( 6 \) using the given probability distribution:

The probabilities are given as:

• \( P(x = 4) = 2k^2 \)
• \( P(x = 5) = 2k \)
• \( P(x = 6) = k^2 + k \)

Therefore, we calculate:

\(P(3 < x \leq 6) = P(x = 4) + P(x = 5) + P(x = 6) = 2k^2 + 2k + (k^2 + k)\)

Simplifying the expression:

\(= 2k^2 + k^2 + 2k + k = 3k^2 + 3k\)

As per the probability axioms, the sum of all probabilities is 1, i.e.,

\(2k + k + 3k + 2k^2 + 2k + k^2 + k + 7k^2 = 1\) \(= 2k + k + 3k + 2k^2 + 2k + k^2 + k + 7k^2 = 1 \Rightarrow 15k^2 + 7k = 1\)

Solving the equation for \( k \), we get:

\(k = \frac{1}{11}\)

Substituting \( k = \frac{1}{11} \) into \( 3k^2 + 3k \), we get:

\(3\left(\frac{1}{11}\right)^2 + 3\left(\frac{1}{11}\right) = 3\left(\frac{1}{121}\right) + \frac{3}{11} = \frac{3}{121} + \frac{33}{121} = \frac{36}{121}\)

Therefore, the probability \( P(3 < x \leq 6) = 0.64 \).

Thus, the correct answer is:

0.64