Problem: You have two options

1: One successful shot at the hoop

2: 2 of 3 successful shots at the hoop

Let P be the probability of making a successful hit, for what value of P will you choose option 2 over 1.

Solution:

If P is probability for option 1

Option 2, we can have 2 cases of success, one all 3 shots are successful – P*P*P

or 2 Hits and 1 Miss (HHM, HMH, MHH) 3 cases, each with probability of P*P*(1-P) i.e. 1-P for one miss. as there are 3 cases, total probability is 3*P*P*(1-p). Combining both result for option 2, P*P*P + 3*P*P*(1-P).

To prefer option 2,

P*P*P + 3*P*P*(1-P)>P

P*P+3P(1-P)>1

Can be solved to quadratic equation

-2P^2+3P-1>0

or

(2P-1)(P-1)>0

P>1 not possible so

P>1/2 or .5

Choose option 2 when probability of hitting hoop is more than 0.5