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.

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.

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:

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

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: