Question

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)

Hint:

Group the outcomes by their common probability value and use \(E[X] = \sum x \cdot \Pr(X=x)\) on each group separately before adding.

Answer 1

Correct Answer: 4.25

Step 1: Verify the probabilities form a valid distribution.
There are 4 values with probability \(\frac{1}{6}\) each (namely 1, 2, 5, 7) and 4 values with probability \(\frac{1}{12}\) each (namely 3, 4, 6, 8). The total probability is \(4 \times \frac{1}{6} + 4 \times \frac{1}{12} = \frac{4}{6} + \frac{4}{12} = \frac{2}{3} + \frac{1}{3} = 1\), so this is a valid probability mass function.

Step 2: Apply the definition of expected value, \(E[X] = \sum_x x \cdot \Pr(X=x)\).
Split the sum into the two probability groups.

Step 3: Sum the values with probability \(\frac{1}{6}\).
\(1 + 2 + 5 + 7 = 15\). Their combined contribution is \(15 \times \frac{1}{6} = \frac{15}{6} = 2.5\).

Step 4: Sum the values with probability \(\frac{1}{12}\).
\(3 + 4 + 6 + 8 = 21\). Their combined contribution is \(21 \times \frac{1}{12} = \frac{21}{12} = 1.75\).

Step 5: Add the two contributions.
\(E[X] = 2.5 + 1.75 = 4.25\).

Step 6: Round to two decimal places as required, which leaves it unchanged at 4.25.

\[ \boxed{4.25} \]