Question

A customer at ShoreBank is called a “frequent ATM user” if he/she uses an ATM for a majority of his/her transactions. According to bank records of transactions, 40% of all ShoreBank customers are frequent ATM users. Anna and Bill are part of a team of market research specialists at ShoreBank that are investigating ways to improve customer services at the bank.

The team has decided to form a customer focus group by choosing 10 ShoreBank customers at random.

(a) Anna is of the opinion that frequent ATM users have the most to contribute to the focus group. She hopes that at least four members of the focus group will be frequent ATM users. What is the probability of this event happening?

(b) Bill, on the other hand, thinks that the input from those customers who are not frequent ATM users is even more important for the focus group. He hopes that the focus group will include at least four customers who are not frequent ATM users. What is the probability of this event happening?

(c) What is the probability that the focus group will satisfy both Anna’s and Bill’s wishes?  

Answer 1

Let X be a Binomial random variable which denotes the number of ShoreBank customers being frequent ATM users

n = 10, p = 0.4

(a) The required probability = P(X ≥ 4)

= 1 - P(X < 4)

= 1 - {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)}

= 1 - {$0.6^{10}&plus;\binom{10}{1}*0.4*0.6^{9}&plus;\binom{10}{2}*0.4^{2}*0.6^{8} &plus; \binom{10}{3}*0.4^{3}*0.6^{7}$}

= 0.6177

(b) The required probability = P(X' ≥ 4) = P(10 - X ≥ 4)

= P(X ≤ 6)

= P(X < 4) + P(X = 4) + P(X = 5) + P(X = 6)

= (1 - 0.6177) + $\binom{10}{4}*0.4^{4}*0.6^{6}&plus;\binom{10}{5}*0.4^{5}*0.6^{5} &plus; \binom{10}{6}*0.4^{6}*0.6^{4}$

= 0.9452

(c) To satisfy both Anna's and Bill's wishes , X should be atleast 4 and (10 - X) should be atleast 4

Thus, X = {4,5,6} satisfies this condition

So, the required probability = P(X = 4) + P(X = 5) + P(X = 6)

= 0.5629