# You roll a die. If it comes up a 6,

You roll a die. If it comes up a 6, you win 100. If not, you gettoroll again. If you get a 6 the second time, you win 50. If not, you lose. Find the expected amount you'll win.
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Provere
Step 1
$\begin{array}{|cccc|}\hline x& 0& 50& 100\\ P\left(X=x\right)& \frac{5}{6}×\frac{5}{6}=\frac{25}{36}& \frac{5}{6}×\frac{1}{6}=\frac{5}{36}& \frac{1}{6}\\ \text{Prob. in decimals}& 0.6944& 0.1389& 0.1667\\ \hline\end{array}$
Step 2
The expected value is the sum of the product of each possibility x with its probability $P\left(X=x\right)$:
$\mu =E\left(X\right)=\sum xP\left(x\right)=0×\frac{25}{36}+50×\frac{5}{36}+100×\frac{1}{6}=\frac{850}{36}\approx \mathrm{}23.61$
Thus the expected amount you will in $\mathrm{}23.61$
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Troy Lesure
Step 1
We have that "x" is the amount of money earned:
We have to fail twice, the probability of failure is $\frac{5}{6}$ and the probability of hitting is $\frac{1}{6}$, therefore if you lose the probability would be:
$\frac{5}{6}×\frac{5}{6}=\frac{25}{36}$
The one to win 50 $, would be: $\frac{1}{6}×\frac{5}{6}=\frac{5}{36}$ and the one to win$ 100 would be:
$\frac{1}{6}+\frac{5}{6}=\frac{6}{36}$
that is to say:
$\begin{array}{|cccc|}\hline x& 0& 50& 100\\ p\left(x\right)& \frac{25}{36}& \frac{5}{36}& \frac{6}{36}\\ \hline\end{array}$