SOLUTION: Suppose that 1,000 tickets are sold for a raffle. If there is one prize of $2,500, two prizes of $500, four prizes of $100, and thirteen prizes of $20, what is the expected value

Question 872564: Suppose that 1,000 tickets are sold for a raffle. If there is one prize of $2,500, two prizes of $500, four prizes of $100, and thirteen prizes of $20, what is the expected value of a raffle ticket? Note: A price for the tickets is not given.
A.$1.91
B.$4.16
C.$5.50
D.$6.76
E. None of these
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Since "A price for the tickets is not given" we'll assume the price is $0 (ie they're free)

let
A: the event you win prize A, prize of $2500
B: the event you win prize B, prize of $500
C: the event you win prize C, prize of $100
D: the event you win prize D, prize of $20
E: the event you do not win any prizes (you receive $0 and don't lose any money).

There are 1+2+4+13 = 20 total prizes (ranging from A to D). Since there are 1000 tickets, this means 1000-20 = 980 tickets won't win any prizes.

P(A) means "probability of winning prize A" and V(A) means "the value or winnings (with respect to the person receiving the money) of prize A". The same applies for events B through D.

E[X] = expected value
E[X] = P(A)*V(A)+P(B)*V(B)+P(C)*V(C)+P(D)*V(D)+P(E)*V(E)
E[X] = (1/1000)*(2500)+(2/1000)*(500)+(4/1000)*(100)+(13/1000)*(20)+(980/1000)*(0)
E[X] = (0.001)*(2500)+(0.002)*(500)+(0.004)*(100)+(0.013)*(20)+(0.980)*(0)
E[X] = 2.5+1+0.4+0.26+0
E[X] = 4.16

The expected value is $4.16
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