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Sagot :
To answer this question, we need to find, first of all, the corresponding probability for the events. Then, we have:
1. The probability of an even number is:
We have that in a single 6-sided die, we have that the only even numbers are 2, 4, and 6. If we roll the die one time, then the probability of this event is:
[tex]P(\text{even)}=\frac{3}{6}[/tex]2. The probability of resulting 1 or 3 is - if the die is rolled one time:
[tex]P(1,3)=\frac{2}{6}[/tex]3. The probability of resulting in a 5 is - if the die is rolled one time:
[tex]P(5)=\frac{1}{6}[/tex]Then, if we add all the corresponding probabilities we have:
[tex]P(\text{total)}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1[/tex]The expected value of the game
To find the expected value of the game, we have to find the product of the probability by the corresponding amount of money of the event as follows:
[tex]E(v)=\frac{3}{6}\cdot-\$5+\frac{2}{6}\cdot\$1+\frac{1}{6}\cdot\$8[/tex][tex]E(v)=-\$2.5+\$(\frac{1}{3})+\$(\frac{4}{3})=-\$2.5+\$(\frac{5}{3})=-\$(\frac{5}{6})=-\$0.833333333333[/tex]Or
[tex]E(v)=-\$0.833333333333[/tex]If we round the answer in terms of dollars rounded to the nearest cent (hundredth), we have that the expected value is:
[tex]E(v)=-\$0.83[/tex]In other words, if we play the game, we will expect to lose 83 cents of a dollar (per game) or 0.83 dollars.
In summary, we have that the expected value of the game is -$0.83.
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