A Bank believes that 83% of its customers prefer the bank to close early on Friday but be open on Saturday morning. Suppose 8 customers are randomly surveyed, and that the binomial distribution applies.

(a) What is the probability that exactly 3 of the surveyed customers like the Saturday option?

(b) What is the probability that exactly 5 of these customers like the Saturday option?

(c) What is the probability that at most 5 customers in the sample like the Saturday option?

(d) What is the probability that at least 5 of them like the Saturday option?

Answer :

Given :

Probability of success $(p)=0.83$

Sample size $(n)=8$

Solution :

A) The probability that exactly 3 of the surveyed customers like the Saturday option :

$$P(x=3)=(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=(\genfrac{}{}{0ex}{}{8}{3})\cdot {0.83}^3\cdot (1–0.83{)}^{8-3}$$ $$=0.0045$$

B) The probability that exactly 5 of these customers like the Saturday option :

$$P(x=5)=(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=(\genfrac{}{}{0ex}{}{8}{5})\cdot {0.83}^5\cdot (1–0.83{)}^{8-5}$$ $$=0.1084$$

C) The probability that at most 5 customers in the sample like the Saturday option :

$$P(x\le 5)=\sum _0^5(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _0^5(\genfrac{}{}{0ex}{}{8}{x})\cdot {0.83}^x\cdot (1–0.83{)}^{8-x}$$ $$=0.1412$$

D) The probability that at least 5 of them like the Saturday option :

$$P(x\ge 5)=\sum _5^8(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _5^8(\genfrac{}{}{0ex}{}{8}{x})\cdot {0.83}^x\cdot (1–0.83{)}^{8-x}$$ $$=0.9672$$