Question

Suppose a basketball player is an excellent free throw shooter and makes 9191​% of his free throws​ (i.e., he has a 9191​% chance of making a single free​ throw). Assume that free throw shots are independent of one another. Suppose this player gets to shoot threethree free throws. Find the probability that he misses all threethree consecutive free throws. Round to the nearest​ ten-thousandth.

Answer 1

0.007

Step-by-step explanation:

Answer 2

`P_(miss)=0.00073`

Step-by-step explanation:

Given Probability of Scoring = 91%

Thus probability of missing = 9%

Probability of missing 3 consecutive throws =
`P_(miss)=P_(miss1)* P_(miss2)* P_(miss3)\\\\P_(miss)=((9)/(100))^(3)\\\\P_(miss)=0.000729`

`P_(miss)=0.0007`