Question

Gerrit attempted 54 free throws and made 27. What is the probability in a two attempt free - throw situation that he will make 0 points ? 1point ? 2 points ?

Answer 1

prob(hit) = 27/54 = 1/2
prob( miss) = 1/2

prob (no points)
= prob(miss, miss) = (1/2)(1/2) = 1/4
prob( 1 point)
= prob(HM) + Prob(MH) = 2(1/4) = 1/2

prob(2 points) ]= prob (HH) = 1/4

(notice 1/4 + 1/2 + 1/4 = 1 , as expected)