A college basketball player makes 80% of his free throws. At the end of a game, his team is losing by two points. He is fouled attempting a three-point shot and is awarded three free throws. Assuming free throw attempts are independent, what is the probability that he makes at least two of the free throws?

.. P(at least 2 of 3 baskets)
= P(2 of 3 baskets) + P(3 of 3 baskets)
= (basket)(basket)(no basket) + (basket)(no basket)(basket) + (no basket)(basket)(basket) + (basket)(basket)(basket)
= (0.8)(0.8)(0.2) + (0.8)(0.2)(0.8) + (0.2)(0.8)(0.8) + (0.8)(0.8)(0.8)
= 3(0.8)(0.8)(0.2) + 0.8^3
= 0.384 + 0.512
= 0.896

QED.

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AFOKE88 AFOKE88 · Answer 2 · 2022-04-15T13:08:57+02:00

The probability that he makes at least two of the free throws from the percentage of success and failure is; 0.896 or 89.6%

What is the probability of selection?

We are told he makes 80% of the throws.

Thus, percentage of failed throws = 20%

Now, we want to find the probability that he makes at least two of the free throws. This is expressed as;

P(at least 2 of 3 baskets) = P(2 of 3 baskets) + P(3 of 3 baskets)

P(2 of 3 baskets) = (0.8 × 0.8 × 0.2) + (0.8 × 0.2 × 0.8) + (0.2 × 0.8 × 0.8) = 0.384

P(3 of 3 baskets) = (0.8 × 0.8 × 0.8) = 0.512

Thus;

P(at least 2 of 3 baskets) = 0.384 + 0.512

P(at least 2 of 3 baskets) = 0.896

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