P(winning) = s(2, 3) + s(3, 3)
The probability of making all the three shots is:s(3, 3) =The probability of making exactly two shots is: P(making 1 and 2, & missing 3) + P(making 1 and 3, & missing 2) + P(making 2 and 3, & missing 1)
= [p * p * (1-p)] + [p * (1-p) * p] + [(1-p) * p * p] = 3*(1-p)*Adding these together, we get:
=Which game you should play? You should play Game 1 if P(Game 1) > P(Game 2):+ 3(1-p) = + 3 - 3 = 3 - 2
p > 3Both terms must be positive, or both must be negative, But we know p < 1, so p – 1 < 0. This means both terms must be negative.- 2 1 > 3p - 2 2 - 3p + 1 > 0 (2p - 1)(p - 1) > 0
2p - 1 < 0 2p < 1 p < 0.5So, we should play Game 1 if 0 < p < 0.5 and Game 2 if 0.5 < p < 1. If p = 0, 0.5, or 1 then P(Game 1) = P(Game 2), so it doesn't matter which game we play.