Question 1162755: Questions are based on the following information: The fine print on an instant lottery ticket claims that one in nine tickets win a prize.
What is the probability that you win at least once if you purchase five tickets?
What is the probability that you win at least twice if you purchase ten tickets?
What is the approximate probability that you win more than 120 times if you purchase 900 tickets?
Of all awarded prizes, 10% are worth Ghc1000, 20% are worth Ghc100, and 70% are worth $10. Find the expected winnings if you purchase a single ticket.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! probability you don't win is 8/9 each time
for 5 times NOT winning at all, the probability is (8/9)^5=0.5549
all other possibilities are winning at least once, and that is 1-0.5549=0.4451
For 10 tickets, probability don't win is (8/9)^10=0.3079
probability win 1 time is 10*(8/9)^9*(1/9), because 10 ways can win once
That probability is 0.3849
Those two sum to 0.6929
All other probabilities are winning at least 2 or more times, and that is 1-0.6929=0.3071
np=900*(1/9)=100 times
variance is np(1-p)=100*(8/9)=88.89
sd is sqrt (V)=9.43
normal approximation is z>(120-100)/9.43 or z>2.12, and that probability is 0.0170.
E(X)=(1/9)(0.1*1000+0.2*100+0.7*10)
=(1/9)(127)
=Ghc14.11; I am ignoring the dollar sign for the last prize.
This is the expected value for the winnings. The overall expected value cannot be determined, since the price of the tickets is not unknown. That would be subtracting (8/9)*price of the ticket from the above figure.
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