Question

2. Consider now the following lottery: There are infinitely many tickets, of which 4% are a “win”. Every ticket costs $2, and a winning ticket pays out $120.

   (a) Give a lower bound on the probability that playing the lottery will win you money. (b) Find a price such that your losses can exceed this price with probability at most 25%.

Answer 1

Given: The probability of winning is, P (W)= 0.04 The cost price of a ticket is, CP= $2 Wining Amount is, WA= $120 (a) A contestant will break even if he/she buys 60 tickets and wins one of them, i.e. the cost of 60 tickets will be ($2 x 60 =) $120 and the winning amount is also $120 so the contestant will win back the amount spent on the tickets. To win money the contestant should buy 59 tickets and win one of them. Then the cost of 59 tickets will be ($2 x 59 =) $118 and the winning amount is $120 so the contestant will win $2 more than the amount spent on the tickets. This amount(i.e. $2) is the lower bound of the winning amount. Compute the probability of winning a ticket among the 59 tickets bought- P(Winning a ticket among the 59 tickets) = P(Winning)*P(Losing) = 0.04x(1-0.04) = 0.0037479 0.00375 Therefore, the lower bound on the probability that playing the lottery will win the contestant money is 0.00375. (b) Let X= lost amount Letn=number of tickets bought. The probability of losing is, P (L) = 1-P (W)= 1 -0.04 = 0.96 It is provided that the losses exceed the cost price of buying n tickets with probability at most 25%, i.e. the lost amount is more than the winning prize of $120 with probability at most 0.25

Compute the number of tickets bought as follows- P(L)" = 0.25 (0.96)* = 0.25 nlog(0.96)=log (0.25) _log(0.25) n= log(0.96) = 33.9595 34 The cost price of buying 34 tickets is: Loss > $120 34x Price of 1 ticket > $120 34x p > $120 p> 3.5294 p> $3.53 Therefore, the price of a ticket is more than $3.53