Question

Let us suppose that each time a basketball player shoots a three-point attempt, the probability of scoring is 0.4. How many three-point attempts does this player need to make for the probability of at least three successes to exceed 0.75?

Answer 1

The probability of 0 successes for n attempts is 0.6^n. For 1 success it is 0.4*0.6^(n-1)*n, for 2 successes it is 0.4^2*0.6^(n-2)*(n choose 2). We want to find when the sum of these is less than 0.25. When n=9 it is 0.23 and when n=8 it is 0.31 so it is 9.