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Sagot :
We have here a case in which we need to find the expected value, and this value is obtained by having the probability of the event times the amount of money (in this case).
We also need to remember that the expected value is a value that we are about to have after making a statistic experiment for a large number of trials.
We can see that the possible outcomes for two dice are 36, and we can see that the probability for double ones, double twos, double threes, and so for is:
[tex]P(doubles)=\frac{6}{36}=\frac{1}{6}[/tex]Now, the probability of not having doubles is:
[tex]P(\text{ not-doubles\rparen}=\frac{30}{36}=\frac{5}{6}[/tex]However, the probability for each double is:
[tex]P(\text{ each-double\rparen}=\frac{1}{36}[/tex]And since we have six possible doubles, then the sum of the probabilities is 6/36 = 1/6.
Now, the expected value here can be found as follows:
[tex]E(X)=\frac{1}{36}(1)+\frac{1}{36}(2)+\frac{1}{36}(3)+\frac{1}{36}(4)+\frac{1}{36}(5)+\frac{1}{36}(6)+\frac{30}{36}(-1)[/tex]We can use -$1 since we lose money besides doubles. Then, the expected value is as follows:
[tex]\begin{gathered} \frac{1}{36}(1+2+3+4+5+6)+\frac{30}{36}(-1) \\ \\ \frac{1}{36}(21)-\frac{30}{36} \\ \\ \frac{21}{36}-\frac{30}{36}=-\frac{9}{36}=-0.25 \end{gathered}[/tex]Then we can expect that we will lose about a quarter of a dollar playing this game.
Therefore, in summary, the expected value of the game is -0.25, or to lose 25 cents if we play this game over and over again, and we will expect to lose money.
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