statistic and probability

Question

Gwen, a friend of Mary from the previous question, is also practicing free throws. However, she is trying to score 3 points in a single set. She will keep shooting sets until she has three successful shots in a single set. Gwen is more confident in her abilities, and believes that she can successfully make any single shot with a probability of 0.8.

Give your answer as a decimal to 4 decimal places.

a) Given the information above, how many sets does Gwen expect to make?

b)b) Given the information above, what is the variance for the number of sets Gwen will make?

c) Given the information above, how many shots does Gwen expect to make?

Answer 1

Hi Ank,

We can assume that each set is independent. now the probability of scoring 3 points is 0.8*0.8*0.8 = 0.512.

thus if she is playing a set, she can get desired success with probability of 0.512.

a)Each set represents an independent binomial trial. So number of trials needed to achieve desired outcome follows binomial distribution.

Now she needs just one success i.e. expected value of randomized variable is 1.

So, we can write - E(X) = Number of trials * probability of success in each trial

thus, 1 = n*0.512

Therefore, n=2.

b) the variance in binomial distribution is given by n*p*(1-p) where P is probability of success.

So, variance = 2*0.5*0.5 i.e. 0.5.

c) Number of shots = No. of shots in 1 set * No of sets played