SOLUTION: You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $410 prize, two $105

Algebra ->  Probability-and-statistics -> SOLUTION: You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $410 prize, two $105      Log On


   

Question 1086168: You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $410 prize, two $105 prizes, and four $30 prizes. Find your expected gain or loss.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A = event of winning the $410 prize
B = event of winning the $105 prize
C = event of winning the $30 prize
D = event of winning no prize (winning $0)

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Probabilities for each event

P(A) = 1/100 = 0.01
P(B) = 2/100 = 0.02
P(C) = 4/100 = 0.04
P(D) = 1-P(A)-P(B)-P(C)
P(D) = 1-0.01-0.02-0.04
P(D) = 0.93

Another way to think of P(D) is to do

P(D) = (number of ways to get no winning tickets)/(number of tickets total)
P(D) = 93/100
P(D) = 0.93

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Net Value (to the ticket holder) for each event

V(A) = net value if event A happens
V(A) = (winnings for event A) - (cost of ticket)
V(A) = 410 - 10
V(A) = 400

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V(B) = net value if event B happens
V(B) = (winnings for event B) - (cost of ticket)
V(B) = 105 - 10
V(B) = 95

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V(C) = net value if event C happens
V(C) = (winnings for event C) - (cost of ticket)
V(C) = 30 - 10
V(C) = 20

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V(D) = net value if event D happens
V(D) = (winnings for event D) - (cost of ticket)
V(D) = 0 - 10
V(D) = -10

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E = Expected Value
E = Sum[P(X)*V(X)] where X is some event
E = P(A)*V(A)+P(B)*V(B)+P(C)*V(C)+P(D)*V(D)
E = 0.01*400 + 0.02*95 + 0.04*20 + 0.93*(-10)
E = -2.60

The expected value is -2.60, which means that on average, you expect to lose $2.60 for each game played.