Question

Tim Duncan is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 72% of the time. What is the probability of Tim Duncan making all of his next 8 free throw attempts?

Answer 1

Answer:

Hence, the probability of Tim Duncan making all of his next 8 free throw attempts is:

$(\dfrac{18}{25})^8=0.0722204136$

Step-by-step explanation:

As the probability of one free throw does not depend on the any other i.e. the probability of each throw is independent.

Hence, the probability of making all of his next 8 free throw is calculated as:

Probability of next 8 free throw=Probability of first free throw×probability o f second free throw×··················× Probability of 8th free throw.

i.e. it is given by:

$Probability=\dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\times \dfrac{72}{100}\\\\\\Probability=(\dfrac{72}{100})^8\\\\Probability=(\dfrac{18}{25})^8$

Hence, the probability is:

$(\dfrac{18}{25})^8=0.0722204136$

Answer 2

We have that the he probability of Tim Duncan making all of his next 8 free throw attempts  is mathematically given as

$P=0.72^8$

From the question we are told

• Tim Duncan is shooting free throws.
• Making or missing free throws doesn't change the probability that
• he will make his next one, and he makes his free throws 72% of the time.
• What is the probability of Tim Duncan making all of his next 8 free throw attempts?

Arithmetic

Generally the equation for the  binomial probability is mathematically given as

$P=\frac{n!}{X!(n-X)!}*p^X*(1-p)^{n-X}\\\\Therefore\\\\ P=\frac{8!}{8!(8-8)!}*0.72^8*(1-0.72)^{0}$

$P=0.72^8$

For more information on Arithmetic visit

https://brainly.com/question/22568180