Respuesta :
Answer:
The expected value of the profit is $-1
Step-by-step explanation:
The expected value of any discrete variable X is calculate as:
E(X)=X1*P(X1)+X2*P(X3)+...+X3*P(X3)
Where X1, X2, ..., X3 are the values that the variable can take and P(X1), P(X2), ..., P(X3) are their probabilities.
In this case the variable X is the dollars that can win, so:
X1=$2,000
X2=$300
x3=$20
X4=$0
Then the probabilities can be calculate as:
[tex]P(X1)=\frac{1}{800}[/tex]
[tex]P(X2)=\frac{3}{800}[/tex]
[tex]P(X3)=\frac{15}{800}[/tex]
[tex]P(X4)=\frac{781}{800}[/tex]
Replacing the variables and probabilities on the equation of expected value we get:
[tex]E(X)=(2000*\frac{1}{800} )+(300*\frac{3}{800} )+(20*\frac{15}{800} )+(0*\frac{781}{800} )[/tex]
E(X)=$4
Additionally, the student bought a ticket by $5, so the expected profit can be calculate as:
Expected Profit = Expected Earnings - Cost = $4 - $5 = $-1
Finally, the expected value of the profit is $-1
