Problem Statement:
A box contains 5 balls labeled from 1 to 5. If three balls are drawn with replacement, let X be the number of distinct balls that are drawn at least once in the three draws. Find E(X).
Answer:
\(P(X=3)=\frac{5}{5}\times\frac{4}{5}\times\frac{3}{5}=\frac{12}{25}\)
\(P(X=2)=\frac{5}{5}\times\frac{1}{5}\times\frac{4}{5}\times\text{C}^3_2=\frac{12}{25}\)
\(P(X=1)=\frac{5}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}\)
\(E(X)=3 \times P(X=3)+2 \times P(X=2)+1 \times P(X=1)\)
\(=3\times\frac{12}{25}+2\times\frac{12}{25}+1\times\frac{1}{25}\)
\(=\frac{61}{25}\)
There are approximately 2.44 distinct balls on average when drawing 3 times with replacement from 5 distinct balls.
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