Question

Chauncey Billups, a current shooting guard for the Los Angeles Clippers, has a career free-throw percentage of 89.4%. Suppose he shoots six free throws in tonight’s game. What is the probability that Billups makes all six free throws?

Answer 1

Answer: 0.5105

Step-by-step explanation:

We know that if events A and B are independent, then the probability of getting A and B is

P(A and B)= P(A) x P(B)  [Product of the probabilities of each event ]

Given :  Chauncey Billups, a current shooting guard for the Los Angeles Clippers, has a career free-throw percentage of 89.4%.

i.e. P( make free-throw)= 0.894

Since all the shots are independent.

Then , the probability that Billups makes all six free throws will be :

$(0.894)\times (0.894)\times (0.894)\times (0.894)\times (0.894)\times (0.894)=(0.894)^6=0.510534520424\approx0.5105$

Hence, the probability that Billups makes all six free throws= 0.5105