GATE 2026 CS (CS1) – Question 58
Let $X$ be a random variable which takes values in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$. Further, $\Pr(X = 1) = \Pr(X = 2) = \Pr(X = 5) = \Pr(X = 7) = \frac{1}{6}$ and $\Pr(X = 3) = \Pr(X = 4) = \Pr(X = 6) = \Pr(X = 8) = \frac{1}{12}$. The expected value of $X$, denoted by $E[X]$, is equal to ___________. (rounded off to two decimal places)
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Correct answer: 4.25
Explanation
First, verify total probability:
$$\sum_{x=1}^8 \Pr(X = x) = 4 \times \frac{1}{6} + 4 \times \frac{1}{12} = \frac{2}{3} + \frac{1}{3} = 1$$
The expected value $E[X]$ is given by:
$$E[X] = \sum_{x=1}^8 x \cdot \Pr(X = x)$$
Split the sum according to the two probability groups:
$$E[X] = \frac{1}{6}(1 + 2 + 5 + 7) + \frac{1}{12}(3 + 4 + 6 + 8)$$
Calculate each partial sum:
- $1 + 2 + 5 + 7 = 15$
- $3 + 4 + 6 + 8 = 21$
Substitute the sums:
$$E[X] = \frac{15}{6} + \frac{21}{12} = \frac{30}{12} + \frac{21}{12} = \frac{51}{12} = \frac{17}{4} = 4.25$$
The correct answer is 4.25.