91Ó°ÊÓ

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

We are given,

Fertile, female cats produce an average of three litters per year and one is chosen randomly from the lot.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of litters a fertile female cat can produce.

The list of values of X are$0,1,2,3,.....$

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$X~P\left(3\right)$

The probability that she has no litters in one year is:

$$\begin{gathered} P(X=0)=\frac{3^0}{0!}{e}^{-3} \\ P(X=0)=0.0498 \end{gathered}$$

The probability that she has at least two litters in one year is:

$$\begin{gathered} P_n(\mathrm{ξ}â‰\yen 2)=P_n\left(0\right)+P_n\left(1\right)+P_n\left(2\right) \\ {P}_n(\mathrm{ξ}â‰\yen 2)=0.00335+0.01907+0.05436 \\ {P}_n(\mathrm{ξ}â‰\yen 2)=0.8009 \end{gathered}$$

The probability that she has exactly three litters in one year is:

$$\begin{gathered} P_n(\mathrm{ξ}=3)=P_n\left(0\right)+P_n\left(1\right)+P_n\left(2\right)+P_n\left(3\right) \\ {P}_n(\mathrm{ξ}=3)=0.00335+0.01907+0.05436+0.01327 \\ {P}_n(\mathrm{ξ}=3)=0.2240 \end{gathered}$$