91Ó°ÊÓ

We have given Poisson rate is $\mathrm{λ}$.

We need to find the probability that both arrived during the first 20 minutes.

The conditional probability of $N(1/3)=2$, $N\left(1\right)=2$. It is equivalent to $N(1/3)=2,N(2/3)=0$, so by independence, we have

$$\begin{gathered} P\left(N\right(1/3)=2\overline{)N\left(1\right)=2)=\frac{P\left(N\right(1/3)=2,N(1)=2)}{P\left(N\right(1)=2)}} \\ =\frac{P\left(N\right(1/3)=2,N(2/3)=0)}{P\left(N\right(1)=2)} \\ =\frac{{\left({\displaystyle \frac{\mathrm{λ}}{3}}\right)}^2\frac{1}{2!}{e}^{-{\displaystyle \frac{\mathrm{λ}}{3}}}.{\left({\displaystyle \frac{2\mathrm{λ}}{3}}\right)}^0\frac{1}{0!}{e}^{-{\displaystyle \frac{2\mathrm{λ}}{3}}}}{{\mathrm{λ}}^2{\displaystyle \frac{1}{2!}}{e}^{-\mathrm{λ}}}=\frac{1}{9} \end{gathered}$$

We have given Poisson rate is $\mathrm{λ}$.

We need to find the probability that at least one arrived during the first 20 minutes.

Similarly from the above part, we have that

$$\begin{gathered} P\left(N\right(1/3)=0\overline{)N\left(1\right)=2)=\frac{P\left(N\right(1/3)=0,N(1)=2)}{P\left(N\right(1)=2)}} \\ =\frac{P\left(N\right(1/3)=0,N(2/3)=2)}{P\left(N\right(1)=2)} \\ =\frac{{\left({\displaystyle \frac{\mathrm{λ}}{3}}\right)}^0{\displaystyle \frac{1}{0!}}{e}^{-{\displaystyle \frac{\mathrm{λ}}{3}}}.{\left({\displaystyle \frac{2\mathrm{λ}}{3}}\right)}^2{\displaystyle \frac{1}{2!}}{e}^{-{\displaystyle \frac{2\mathrm{λ}}{3}}}}{{\mathrm{λ}}^2{\displaystyle \frac{1}{2!}}{e}^{-\mathrm{λ}}}=\frac{4}{9} \end{gathered}$$

So, the conditional probability that some person arrived during the first$20$minutes in which $2$people has arrived within the first hour is$\frac{5}{9}$.