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

We are given a discrete random variable \(X\) taking values in \(\{1,2,3,4,5,6,7,8\}\) with two distinct probability values assigned to two groups of outcomes.

Step 1: Write down the given probabilities.

\(Pr(X=1)=Pr(X=2)=Pr(X=5)=Pr(X=7)=\dfrac{1}{6}\)

\(Pr(X=3)=Pr(X=4)=Pr(X=6)=Pr(X=8)=\dfrac{1}{12}\)

Step 2: Verify that the probabilities form a valid distribution (they should sum to 1).

\[4\times\frac{1}{6}+4\times\frac{1}{12}=\frac{4}{6}+\frac{4}{12}=\frac{2}{3}+\frac{1}{3}=1\]

Since the total probability is 1, the distribution is valid.

Step 3: Recall the formula for expected value of a discrete random variable.

\[E[X]=\sum_{i} x_i \cdot Pr(X=x_i)\]

Step 4: Split the sum into the two groups based on their probability values.

\[E[X]=\frac{1}{6}(1+2+5+7)+\frac{1}{12}(3+4+6+8)\]

Step 5: Compute each group sum.

Group with probability \(\frac{1}{6}\): \(1+2+5+7=15\), so contribution \(=\dfrac{15}{6}=2.5\)

Group with probability \(\frac{1}{12}\): \(3+4+6+8=21\), so contribution \(=\dfrac{21}{12}=1.75\)

Step 6: Add the two contributions to get the final expected value.

\[E[X]=2.5+1.75=4.25\]

This lies in the accepted range \(4.24\) to \(4.26\).

Final Answer:

\[\boxed{E[X]=4.25}\]