Question

John's average for making free throws in a basketball game is .80. In a one-and-one free throw situation (where he shoots a second basket only if he makes the first), what is the probability that he makes exactly one basket?

Answer 1

The probability of John making the first and missing the second is:

`P(makes\ 1\ basket)=0.8*(1-0.8)=0.16`

Answer 2

0.16

Explanation:

Given that John's average for making free throws in a basketball game is .80.

He can get a chance for II throw only if he hits the first

So required probability = Prob that he hits the first but misses the second

Prob for hitting a single trial = 0.8 which is constant for each throw

Prob for missing a single trial = 1-0.8 = 0.2 since there are only two outcomes

Hence required probability = 0.8(1-0.2) = 0.16