# Solved:Items are packed into boxes of 1000, and each item has a probability of 0.001 of having some type of fault

#### ByDr. Raju Chaudhari

Aug 23, 2020

a. Items are packed into boxes of 1000, and each item has a probability of 0.001 of having some type of fault. What is the probability that a box will contain fewer than three defective items?
b. If the company sells 100 000 boxes per year and guarantees fewer than three defectives per box, what is the expected number of guarantee claims?
c. Replacement of a box returned under the guarantee costs £ 150. What is the expected cost of guarantee claims?
d. Boxes sell at £ 100 but cost £60 to produce and distribute. What is the company's expected profit for sales of boxes?

#### Solution

Let $X$ denote the number of defective per box. Each item has a probability of 0.001 of having some type of fault,i.e., $p =0.001$.

Given that $X\sim B(1000, 0.001)$. But as $n$ is too large and $p$ is too small and $\lambda=n*p = 1000*0.001 = 1$.

Thus $X\sim P(\lambda = 1)$

The pmf of Poisson distribution with $\lambda =1$ is

\begin{aligned} P(X=x) &= \frac{e^{-1}(1)^x}{x!},\\ & \qquad x=0,1,2,\cdots \end{aligned}

(a) The probability that a box will contain fewer than three defective items is

The probability that it will receive only two complaints on any given day is
\begin{aligned} P(X < 3) &= P(X\leq 2)\\ & = P(X=0)+P(X=1)+P(X=2)\\ &= \frac{e^{-1}(1)^0}{0!}+\frac{e^{-1}(1)^1}{1!}+\frac{e^{-1}1^2}{2!}\\ &= 0.3679+0.3679+0.1839\\ &= 0.9197. \end{aligned}

(b) If the company sells 100 000 boxes per year and guarantees fewer than three defectives per box, then the expected number of guarantee claims is

\begin{aligned} 100000*P(X < 3) & = 100000*0.9197\\ &=91970 \end{aligned}

(c) Replacement of a box returned under the guarantee costs £ 150, the expected cost of guarantee claims is

\begin{aligned} 150*100000*P(X < 3) & = 150*100000*0.9197\\ &= 13795500 \end{aligned}

(d) Boxes sell at £ 100 but cost £60 to produce and distribute.
The expected sell is

\begin{aligned} \text{Expected sell} &=100*100000*P(X < 3)\\ & = 100*100000*0.9197\\ &=9197000 \end{aligned}

The expected cost is
\begin{aligned} \text{Expected sell} &=60*100000*P(X < 3)\\ & = 60*100000*0.9197\\ &=5518200 \end{aligned}

The company's expected profit = Expected sell - expected cost.

That is 9197000 -5518200 =£ 3678800.