Question

If the random variable \( X \) takes the values 1, 2, 3, and 4 such that \( 2P(X=1) = 3P(X=2) = P(X=3) = 5P(X=4) \), then the mean of \( X \) is

• A. \( \frac{30}{61} \)
• B. \( \frac{90}{61} \)
• C. \( \frac{139}{61} \)
• D. \( \frac{149}{61} \)

Hint:

Use a common variable \( k \) to express probabilities and normalize using the total probability condition.

Answer 1

The Correct Option is D

Let \( P(X = 1) = \frac{k}{2}, \ P(X = 2) = \frac{k}{3}, \ P(X = 3) = k, \ P(X = 4) = \frac{k}{5} \)
Sum of probabilities:
\[ \frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = k \left( \frac{1}{2} + \frac{1}{3} + 1 + \frac{1}{5} \right) = k \cdot \frac{61}{30} \]
Set total probability to 1:
\[ k \cdot \frac{61}{30} = 1 \Rightarrow k = \frac{30}{61} \]
Now compute mean \( E(X) = \sum x_i P(x_i) \):
\[ E(X) = 1 \cdot \frac{30}{61 \cdot 2} + 2 \cdot \frac{30}{61 \cdot 3} + 3 \cdot \frac{30}{61} + 4 \cdot \frac{30}{61 \cdot 5} \]
\[ = \frac{30}{61} \left( \frac{1}{2} + \frac{4}{3} + 3 + \frac{4}{5} \right) = \frac{30}{61} \cdot \left( \frac{15 + 40 + 180 + 24}{30} \right) = \frac{30}{61} \cdot \frac{149}{30} = \frac{149}{61} \]