A basketball player makes 80% of his free throws. We set him on the free throw line and told him to shoot free throws until he misses. Let the random variable X be the number of free throws taken by the player until he misses. Assuming that his shots are independent, what is the probability of this player missing a free throw for the first time on the fifth attempt

0.08192

Step-by-step explanation:

Given: Probability of free throws = 0.80

Then, probability of missing the free throw = 1-0.80=0.20

Now, According to the negative binomial probability distribution, the probability that the rth success is achieved in xth trial is given by ;-

Put p = 0.20 , x=5 and r =1, we get

The probability of this player missing a free throw for the first time on the fifth attempt as

hence the required probability =0.08192

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8