91Ó°ÊÓ

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

We are given,

A bag of $144$fortune cookies and the chance of having an extra fortune in a fortune cookie is $3\%$. Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of fortune cookies that have an extra fortune.

The list of values that we use for values of X are$0,1,2,3,.....,144.$

Here, the random variable X refers to the number of trials. Each trial is independent of others and has similar probability of success. This implies that random variable X follows Poisson Distribution.

The Poisson distribution is shown as$X~P(x,\mathrm{μ})$where x is the successes occurs in poisson distribution and$\mathrm{μ}$is the mean successes.

Thus, the distribution of X is role="math" localid="1649244224297" $X~P(4.32)$

We are given, that in a bag of $144$fortune cookies there is a $0.03$chance of having an extra fortune cookie.

The expected number of cookies that have an extra fortune is calculated as follow:

$$\begin{gathered} X=144×0.03 \\ X=4.32 \end{gathered}$$

The probability in poisson distribution is as follow:

$f\left(x\right)=\frac{{\mathrm{λ}}_x}{x!}{e}^{-\mathrm{λ}}$

where, $\mathrm{λ}=4.32,x=0$

localid="1649246766111" $$\begin{gathered} P_n(X=0)=\frac{4.{32}^0}{0!}{e}^{-4.32} \\ {P}_n(X=0)=0.0124or0.0133 \end{gathered}$$

The probability that more than three have an extra fortune is calculated as follow:

$$\begin{gathered} P_n(\mathrm{ξ}≤3)=P_n\left(0\right)+P_n\left(1\right)+P_n\left(2\right)+P_n\left(3\right) \\ {P}_n(\mathrm{ξ}≤3)=0.00335+0.01907+0.05436+0.01327 \\ {P}_n(\mathrm{ξ}≤3)=0.6300or0.6264 \end{gathered}$$

As n increases the probabilities comes closer together.