GIVEN:
We are given that in a game of chance, drawing a ball with a particular number (between 1 and 8) has particular winning prizes attached to them.
The winning prize for each number drawn is given as indicated.
Required;
To find the expected value of playing this game.
Step-by-step solution;
W calculate the expected value by multiplying the value of each outcome by its probability of occurring and then, sum up all the results.
The probability distribution is as follows;
[tex](\text{\$1}=\frac{1}{8}),\text{ \lparen\$}2=\frac{1}{8}),(\text{\$}5=\frac{1}{8}),(\text{\$}6=\frac{1}{8}),(\text{\$}8=\frac{1}{8}),(\text{\$}10=\frac{1}{8}),(-\text{\$}14=\frac{1}{4})[/tex]Take note of the following;
The probability of choosing each number is 1/8 since there are 8 balls numbered 1 to 8. The probability of choosing 7 or 8 is the addition of the probability of choosing 7 and the probability of choosing 8. The prize for choosing 7 or 8 is a loss and therefore, its a negative value (-$14).
The expected value therefore is;
[tex]EV=[1\times\frac{1}{8}]+[2\times\frac{1}{8}]+[5\times\frac{1}{8}]+[6\times\frac{1}{8}]+[8\times\frac{1}{8}]+[10\times\frac{1}{8}]-[14\times\frac{1}{4}][/tex][tex]\begin{gathered} EV=\frac{1}{8}+\frac{1}{4}+\frac{5}{8}+\frac{3}{4}+1+1\frac{1}{4}-3\frac{1}{2} \\ \\ EV=\frac{1+2+5+6+8+10-28}{8} \\ \\ EV=\frac{4}{8} \\ \\ EV=0.5 \end{gathered}[/tex]The expected value of playing the game is $0.50
If Dan plays the game many times, in the long run he can expect to gain money. The total probabilities of winning when summed up equals
[tex]\begin{gathered} Win=\frac{1}{8}+\frac{1}{4}+\frac{5}{8}+\frac{3}{4}+1+1\frac{1}{4}=\frac{32}{8} \\ \\ Win=4 \end{gathered}[/tex]The probability of losing is
[tex]\begin{gathered} Lose=14\times\frac{1}{4} \\ \\ Lose=3.50 \end{gathered}[/tex]Therefore, Dan can expect to gain money
He can expect to win $4.00 per selection.
Therefore,
ANSWER:
(A) $0.50
(B) Dan can expect to gain money
He can expect to win $4.00 per selection.
