A game at the state fair has a "guesser" guess the month of your birth. You win a prize if he misses your month by more than 2 months (for ex: if you were born in April, and he guesses February, he wins; if he guessed January, you win). For simplicity, assume that all 12 months are equally likely birth months, and that the results of all games are independent.
a. In the course of a night, the guesser has 6 players. What is the distribution of the number of players who win?
b. If the guesser has 6 players, find the probabilty that he wins more than half of those games.
Please help.. I have no idea what I'm doing T_T
Statistics - Reiny, Tuesday, July 3, 2012 at 7:41am
For any given birth-month the guesser wins by guessing the 5 months "around" that month.
e.g. if born in June, the correct choices would be
April, May, June, July , and August, which is 5 months.
Prob(guesser wins) = 5/12
prob(1 win out of 6) = C(6,1) (5/12) (7/12)^5 = ....
prob(2 wins out of 6) = C(6,2) (5/12)^2 (7/12)^4 = ..
prob(3 wins out of 6) = C(6,3) (5/12)^3 (7/12)^3 = ...
prob(6 wins out of 6) = C(6,6)(5/12)^6 = ...
b) prob (he wins more than half of the 6 games)
= (add up the prob of 4, 5, and 6 wins)