Jenny Albright knows that in the long term she has a 60 chan

Jenny Albright knows that in the long term she has a 60% chance of making a sale when calling on a customer. One morning she arranges six calls. a) What is her probability of making exactly three sales? b) What are her probabilities of making other numbers of sales? c) What is her probability of making fewer than three sales?

Solution

This is a binomial distribution problem

P(X = r) = nCr * p^r * (1-p)^n-r

a)

p = probability of making a sale = 60% = 0.6

P(X=3) = 6C3 * 0.6^3 * 0.4^3 = 0.27648

b)

P(X=0) = 6C0 * 0.6^0 * 0.4^6 = 0.004096

P(X=1) = 6C1 * 0.6 * 0.4^5 = 0.036864

P(X=2) = 6C2 * 0.6^2 * 0.4^4 = 0.13824

P(X=4) = 6C4 * 0.6^4 * 0.4^2 = 0.31104

P(X=5) = 6C5 * 0.6^5 * 0.4 = 0.186624

P(X=6) = 6C6 * 0.6^6 = 0.046656

c)

P(X<3) = P(X=0) + P(X=1) + P(X=2)

= 0.1792