Question

Let \(X\) take values in \(\{1,2,3,4,5,6,7,8\}\).
\[\Pr(X=1)=\Pr(X=2)=\Pr(X=5)=\Pr(X=7)=\frac16,\]
\[\Pr(X=3)=\Pr(X=4)=\Pr(X=6)=\Pr(X=8)=\frac1{12}.\]
The expected value \(E[X]\) is \(\underline{\hspace{1cm}}\) (rounded off to two decimal places).

Hint:

Split the sample space into two groups by probability value: \(\{1,2,5,7\}\) each with probability \(1/6\), and \(\{3,4,6,8\}\) each with probability \(1/12\). Then apply \(E[X]=\sum x_i Pr(X=x_i)\) group-wise and add the results.

Answer 1

Correct Answer: 4.24

An alternative way to solve this is by grouping the outcomes according to their probability mass and using a weighted-average style computation instead of expanding term by term.

Step 1: Identify the two probability classes.

Class A = \(\{1,2,5,7\}\), each with probability \(p_1 = 1/6\).

Class B = \(\{3,4,6,8\}\), each with probability \(p_2 = 1/12\). Note that \(p_2 = p_1/2\), so Class B values count 'half as much' as Class A values in the expectation sum.

Step 2: Find the sum of values in each class.

Sum of Class A values: \(S_A = 1+2+5+7 = 15\).

Sum of Class B values: \(S_B = 3+4+6+8 = 21\).

Step 3: Express \(E[X]\) using the class sums directly.

\(E[X] = p_1 \cdot S_A + p_2 \cdot S_B = \frac{1}{6}(15) + \frac{1}{12}(21)\)

Step 4: Convert to a common denominator of 12 for a clean single fraction.

\(\frac{1}{6}(15) = \frac{2}{12}(15) = \frac{30}{12}\), and \(\frac{1}{12}(21) = \frac{21}{12}\).

So \(E[X] = \frac{30 + 21}{12} = \frac{51}{12}\).

Step 5: Simplify the fraction to a decimal.

\(\frac{51}{12} = 4.25\)

As a sanity check, since all 8 probabilities must sum to 1, and they do (as shown by \(4(1/6) + 4(1/12) = 1\)), this confirms our distribution and hence our expectation calculation is consistent.

Therefore \(E[X] = 4.25\), which falls within the marking range of 4.24 to 4.26.

Final Answer: \(E[X] = 4.25\)