Question

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?

Answer 1

.. 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.

Answer 2

Answer:

0.64

Step-by-step explanation:

He makes two from three throws, so he makes three throws or two throws.

We add the probabilities:

0.2*0.8^2+0.8^3

then this is equal to 0.64.