Answer:
Step-by-step explanation:
Given that when a basketball player shoots a free​ throw, the odds in favor of his making it are 2020 to 99.
Prob that he shoots a free throw = [tex]\frac{99}{2020+99} =0.04672[/tex]
a) Probability that when this basketball player shoots a free​ throw, he misses it=1-0.04672 = 0.95328
b) When no of throws = 100 average of throws he makes is
Since x no of throws he makes has two outcomes and each trial is independent , X is binomial with n =100 and p = 0.04672
Expected value of throws = np = 4.672
Answer:
a. 4.672
b. Â 0.95328
Step-by-step explanation:
It is computed that when a basketball player shoots a free​ throw, the odds in favor of his making it are 2020 to 99. Find the probability that when this basketball player shoots a free​ throw, he misses it. Out of every 100 free throws he​ attempts, on the average how many should he​ make?
Probability is the likelihood of an event to occur or not. Probability that an event will occur or not is usually less than 1
when the player shoots a free​ throw, the odds on his side is 2020 to 99.
P(A), probability of shooting a throw = 99/(2020+99)
P(A)=0.04672
a) Probability that when he shoots a free throw, he misses it will be
P(B)=(1-P(A))=1-0.04672 = 0.95328
b) if the number of throws are 100 average of throws he makes is
the no of throws, n=100
probability of having a free throw=0.04672
Expected value of throws = np = 4.672
