# Probability problem with combination of poisson and binomial distributions

Exercise

The number of clients that enter to a bank is a Poisson process of parameter $\lambda>0$ persons per hour. Each client has probability $p$ of being a man and $1-p$ of being a woman. After getting into the bank, each client takes a number in order of arrival and waits to be assisted by one of the two counters available. The first counter assists those clients with an odd number, and the second counter those with an even number. The bank is open $5$ hours per day during the five working days of the week.

1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours.

2) Calculate the probability of at least one client entering to the bank during the day but all clients of the day having entered past the two first hours in at least $3$ days of the week.

3) Knowing that at least a woman gets into the bank, ¿what is the probability of the first woman getting into the bank being assisted by the second counter?

My attempt at a solution

1) I define the random variable $Y_i=\text{number of clients in the i-th hour of the opening hours}$ for $i=1,2,3,4,5$, since $Y_i \sim P(\lambda)$ and these random variables are independent, then $Y=Y_3+Y_4 \sim P(2\lambda)$. The random variable that seems suitable to calculate the probability is $X \sim Bin(8, P(Y=4))$.

We have $P(Y=4)=\dfrac{e^{-2\lambda}(2\lambda)^4}{4!}$ So if $A$ is the event described in 1), then $$P(A)=P(X=4)$$$$={8 \choose 4}(\dfrac{e^{-2\lambda}(2\lambda)^4}{4!})^2(1- \dfrac{e^{-2\lambda}(2\lambda)^4}{4!})^2$$

2) Now I define $Y=Y_3+Y_4+Y_5$, so $Y \sim P(3\lambda)$ , first lets calculate the probability of at least one client entering to the bank in the last three hours:

$$P(Y \geq 1)=1-P(Y=0)$$$$=1-e^{-3\lambda}$$ Now if each day of the working days is considered as a bernoulli experiment in which the success is $Y \geq 1$,then for each of the five days we consider the bernoulli random variable $X_i \sim Ber(P(Y \geq 1))$, the sum of these random variables $X=X_1+...+X_5$ has a binomial distribution, $X \sim Bin(5,P(Y \geq 1))$

The probability in 2) is exactly the probability of $X \geq 3)$

$$P( X \geq 3)= \sum_{i=3}^5 {5 \choose i}(1-e^{-3\lambda})^i(e^{-3\lambda})^{5-i}$$

3) If the first woman is assisted by the second counter, that means that she must have an even number, so the number of men that entered before her must be odd. We can consider a geometric random variable $X \sim \mathcal G(1-p)$, so we want to calculate the probability of $X$ being an even number, $$P(X=2k, k \in N)=\sum_{i=1}^{\infty}p^{2i-1}(1-p)$$$$=\dfrac{1-p}{p}(\sum_{i=0}^{\infty}(p^2)^i-1)$$$$=\dfrac{1-p}{p}(\dfrac{p^2}{1-p^2})=\dfrac{p}{1+p}$$

I would really appreciate if someone could take a look at my solution and correct any mistakes I could possibly have commited. Thanks in advance.

• You can use \left and \right to let the parentheses adjust to their content. – joriki Apr 27 '16 at 20:32

1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours.

I define the random variable $Y_i$ =number of clients in the i-th hour of the opening hours $Y_i$=number of clients in the i-th hour of the opening hours for $i=1,2,3,4$ , since $Y_i \sim P(λ)$ and these random variables are independent, then $Y=Y_3 +Y_4 \sim P(2λ)$ . The random variable that seems suitable to calculate the probability is $X∼Bin(8,P(Y=4))$.

Fine until the last sentence.   Close, but no.

What you have not taken into account is the given condition that $Y_1+Y_2+Y_3+Y_4=8$.   You know that eight Poisson point events are distributed across these time intervals.   But how are they distributed across these intervals?

Uniformly, that's how.   Each of these four events may have occurred in the latter two hours with probability $1/2$, so the count of the events that did so is Binomially distributed, via: $X\sim\mathcal {Bin}(4, 1/2)$

$$\mathsf P(X=4) ~=~ \dbinom{8}{4}\dfrac 1 {2^8} ~=~ \dfrac {35}{128}$$

To verify the argument we work from the premises that (1) the number of clients who arrive in an interval $t$ is Poisson distributed with rate $t\lambda$, and that (2) the count of arrivals in disjoint intervals will be independent.

That is to say: If we let $X_1$ be the count of clients who arrive in the earlier two hours, and $X_2$ be the count of those who do so in the latter two hours.   Then $X_1\sim\mathcal{Pois}(2\lambda)$ and independently $X_2\sim\mathcal {Pois}(2\lambda)$.   Also $\mathcal X_1+X_2\sim\mathcal{Pois}(4\lambda)$

The conditional probability that $k$ clients arrived in the later two hours when given that $n$ arrived somewhen in the four hours is:

\begin{align}\mathsf P(X_2=k\mid X_1+X_2=n) ~=~ &\dfrac{\mathsf P(X_2=k)~\mathsf P(X_1=n-k)}{\mathsf P(X_1+X_2=k)} \\[2ex] ~=~ & \dfrac{\dfrac{(2\lambda)^k~\mathsf e^{-2\lambda}}{k!}~\dfrac{(2\lambda)^{n-k}\mathsf e^{-2\lambda}}{(n-k)!}}{\dfrac{(4\lambda)^n\mathsf e^{-4\lambda}}{n!}} \\[2ex] ~=~ & \dfrac{n!}{k!~(n-k)!}\dfrac 1{2^n} \end{align}

• Oh, let me see if I got it: the number of clients that get into per hour is a random variable with poisson distribution and, for each client, the probability of having arrived at time $t$ for $t \in [0,1]$ is a random variable with uniform distribution, is this correct? – user16924 Apr 28 '16 at 0:45
• @user16924 Yes; it is basically the limit case of the addendumabove. – Graham Kemp Apr 28 '16 at 1:19

1) The conditional distribution of the $8$ clients along the interval $[0,4]$ is Uniform, so the conditional distribution of the number of clients in a sub interval $(2,4]$ is Binomial with $p=\frac{2}{4}$ and $n=8$. Denote it by $X$, the event of interest is $\{X=4\}$, so $P(X=4) = \binom{8}{4}\frac{1}{2^8}$.

The answers to 2 and 3 look OK to me.

• I don't see why the distribution is uniform, isn't the sum of independent random variables with Poisson distribution a random variable with Poisson distribution? – user16924 Apr 28 '16 at 0:13
• see. You can show it analytically by finding the conditional probability mass function of single arrival (From a Poisson process) on some time interval $[0,t)$ – V. Vancak Apr 28 '16 at 0:25
• @user16924 Yes, if $Y_1 \sim\mathcal P(\lambda)~, Y_2\sim\mathcal P(\lambda)$ , and they are independent, then $Y_1+Y_2\sim \mathcal P(2\lambda)$. But the conditional distribution is $$Y_1\mid Y_1+Y_2 \sim \mathcal {Bin}(Y_1+Y_2, \dfrac {1}{2})$$ Because each Point Event known to have occurred is uniformly distributed over the total interval. – Graham Kemp Apr 28 '16 at 0:45