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?

0.16

Step-by-step 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

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step-by-step explanation:

9-15%

step-by-step explanation: