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**Problem:**

You have a basketball hoop and someone says that you can play 1 of 2 games.

Game #1: You get one shot to make the hoop.

Game #2: You get three shots and you have to make 2 of 3 shots.

If p is the probability of making a particular shot, for which values of p should you pick one game or the other?

### Solution:

For game #1, you have probability p of winning.For game #2, you can either make 2 shots of 3, with probability 3p^2(1-p), or make all of the three shots, with probability p^3.

Therefore, to choose game #1, you need to have: p > 3p^2(1-p)+p^3

p > 3p^2 - 2p^3

2p^3 - 3p^2 + p > 0

p(2p^2 - 3p + 1) > 0

p(2p-1)(p-1) & gt;0

here p>0 is obvious as probability cannot be negative. Also, p-1 > 0 is not correct as p cannot be greater than 1. hence p>1/2 is correct.

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