Question

If $X$ is a random variable with p.m.f. as follows.
$$P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, 1 \\ \frac{kx}{48}, & x = 2 \\ \frac{1}{4}, & x = 3 \end{cases}$$
then $E(X) =$

• A. 1.1875
• B. 1.3125
• C. 1.5625
• D. 0.5625

Hint:

Always verify that your calculated $P(X=x)$ values are non-negative and sum to exactly 1 before computing the expectation to catch errors early.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The sum of all probabilities in a probability mass function (p.m.f.) must be equal to 1. We first need to determine the value of constant $k$ by summing the probabilities for all possible values of $X$ ($0, 1, 2, 3$). Once $k$ is found, we calculate the expected value $E(X) = \sum x \cdot P(X=x)$.

Step 2: Key Formula or Approach:
The sum of probabilities is $\sum P(X=x) = 1$. The expected value is $E(X) = \sum_{x=0}^{3} x P(X=x)$.

Step 3: Detailed Explanation:
Given $P(X=0) = 5/16$, $P(X=1) = 5/16$, $P(X=2) = 2k/48 = k/24$, and $P(X=3) = 1/4 = 4/16$.
Summing these: $\frac{5}{16} + \frac{5}{16} + \frac{k}{24} + \frac{4}{16} = 1 \implies \frac{14}{16} + \frac{k}{24} = 1$.
$\frac{k}{24} = 1 - \frac{7}{8} = \frac{1}{8} \implies k = 3$.
Thus, $P(X=2) = \frac{3 \times 2}{48} = \frac{6}{48} = \frac{1}{8} = \frac{2}{16}$.
Now, $E(X) = (0 \cdot \frac{5}{16}) + (1 \cdot \frac{5}{16}) + (2 \cdot \frac{2}{16}) + (3 \cdot \frac{4}{16}) = 0 + \frac{5}{16} + \frac{4}{16} + \frac{12}{16} = \frac{21}{16} = 1.3125$.

Step 4: Final Answer:
The value of $E(X)$ is $1.3125$, which matches option (B).