# Solved (Free) : In a market study for Target, a researcher found that 70% of customers are repeat customers

#### ByDr. Raju Chaudhari

Mar 9, 2021

In a market study for Target, a researcher found that 70% of customers are repeat customers. If 14 customers are selected at random, find the probability of getting,

(a) Exactly 9 of them are repeat customer

(b) At least 12 of them are repeat customers

(c) At most 12 of them are repeat customers.

#### Solution

Let $X$ denote the number of repeat customers.
70% of customers are repeat customers. So $p=0.70$

Given that $n = 14$, $p= 0.7$. $X\sim B(14, 0.7)$.

That is the random variable $X$ follows a Binomial distribution with $n = 14$ and $p= 0.7$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{14}{x} (0.7)^x (1-0.7)^{14-x}, \\ &\quad x=0,1,\cdots, 14 \end{aligned}

(a) The probability that exactly $9$ of them are repeat customers is

 \begin{aligned} P(X = 9) & =\binom{14}{9}(0.7)^{9}(1-0.7)^{14-9}\\\\ & = 0.1963 \end{aligned}

(b) The probability that at least $12$ customers are repeat customers is

 \begin{aligned} P(X\geq 12) & =\sum_{x=12}^{14} P(x)\\\\ & =\sum_{x=12}^{14}\binom{14}{x}(0.7)^x(1-0.7)^{14-x}\\\\ & = (0.11336)+(0.04069)+(0.00678) \\\\ & = 0.16084 \end{aligned}

(c) The probability that at most 12 of them are repeat customers is

 \begin{aligned} P(X\leq 12) & =1-\sum_{x=13}^{14} P(x)\\\\ & =1-\sum_{x=13}^{14}\binom{14}{x}(0.7)^x(1-0.7)^{14-x}\\\\ & = 1-(0.04069)+(0.00678) \\\\ & = 0.95252 \end{aligned}