Question

4. A basketball player practices making 100 free throws every day. The probability that she makes each free throw is p=0.6. You may assume that each free throw's outcome is independent of every other free throw's outcome. a) What is the probability that she makes all 100 free throws? b) What is the probability that she makes half of the 100 free throws? b) What is the expected number of free throws that she misses? c) What is the variance of the number of free throws that she misses?

Answer 1

Answer:

given n=100

p=0.6

A) Let x be number of successful free throws

x=100

So Probability = $_{100}^{100}\textrm{P}*(P^100)*(1-p)^0$

$_{100}^{100}\textrm{P}=100!/100!*(100-100)!$=1

Probability =1* 0.6^100*0.4^0=6.533*e^(-23)

B) She makes half of the throw x=50

So Probability = $_{50}^{100}\textrm{P}*(P^50)*(1-p)^50$

$_{50}^{100}\textrm{P}=100!/50!*(100-50)!$

Probability =100!/(50!*50!)* 0.6^50*0.4^50=0.01033

C)

Expected number throws that misses= 100*0.4=40

D)

Variance of free throws that misses = 100*0.4*(1-0.4)=24