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Suppose you have a box with five balls of different colors. If you draw a ball 100 times and replace it, what is the expected number of different colors you would have after 100 trials?

rezzz
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  • It would be good if you indicated what you had tried. That is currently the usual expectation on MSE. – André Nicolas Jul 12 '15 at 02:12
  • Thank you Andre! suppose that we only drew out two balls, so the expected number of colors is: The expected value is 1(1/5) + 2(4/5) = 9/5 = 1.8

    but I could not generalize that for n.

     – rezzz Jul 12 '15 at 13:18
  • Even this much is helpful. Please remember for the next time. – André Nicolas Jul 12 '15 at 14:52

1 Answers1

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For $i=1$ to $5$, define random variable $X_i$ by $X_i=1$ if colour $i$ is drawn at least once in the $100$ trials, and by $X_i=0$ otherwise. Then the number $Y$ of different colours drawn is given by $Y=X_1+\cdots+X_5$.

By the linearity of expectation we have $E(Y)=E(X_1)+\cdots +E(X_5)=5E(X_1)$.

But $E(X_1)=\Pr(X_1=1)$. And $\Pr(X_1=1)$ is $1$ minus the probability colour $1$ is not drawn. So $E(X_1)=1-(4/5)^{100}$.

André Nicolas
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