Question

A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.

Short answer

Using Poisson Distribution and Binomial distribution of the probability comes out to be $0.8153$and $0.8179$respectively.

Step-by-step solution

Content Introduction

The binomial distribution determines the probability of looking at a specific quantity of a hit results in a specific quantity of trials

In a large population, the Poisson distribution is used to characterize the distribution of unusual events.

 Explanation of Poisson Distribution

First let us use Poisson Distribution,

we are given, $3\%$ of its bulbs are defective and a string of light contains$100$bulbs.

So the formula used here will be:

$$\begin{gathered} \lambda =np \\ \lambda =100(0.03) \\ \lambda =3 \end{gathered}$$

The distribution comes out to be $3$

Now, the probability that a string of 100 lights contains at most four defective bulbs is:

$$\begin{gathered} P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ P(X\le 4)=\frac{3^0}{0!}{e}^{-3}+\frac{3^1}{1!}{e}^{-3}+\frac{3^2}{2!}{e}^{-3}+\frac{3^3}{3!}{e}^{-3}+\frac{3^4}{4!}{e}^{-3} \\ P(X\le 4)=0.8153 \end{gathered}$$

Explanation of Binomial Distribution

Using Binomial distribution , we are given, $3\%$of its bulbs are defective and a string of light contains $100$bulbs. The distribution comes out to be:

$P~B(100,0.03)$where $n=100,p=0.03$

The probability that a string of 100 lights contains at most four defective bulbs is:

$$\begin{gathered} P(x\le 4){=}_{100}{C}_1{(0.3)}^4{(0.7)}^3{+}_{100}{C}_2{(0.3)}^3{(0.7)}^2{+}_{100}{C}_1{(0.3)}^2{(0.7)}^1{+}_{100}{C}_0{(0.3)}^1 \\ P(x\le 4)=0.8179 \end{gathered}$$