Over a weekend of 2 days, it is given that 3 requests are received. Find the expected total income for the weekend.

Question

A small hire company has 2 buses. Each bus can be hired for one whole day at a time. The rental charge per day is RM 600 per bus. The number of request to hire a bus for one whole day may be modelled by a Poisson distribution with mean 1.2. Over a weekend of 2 days, it is given that 3 requests are received. Find the expected total income for the weekend.

Answer 1

We know that in the two days three requests were received. There are four possibilities for this to happen, namely $$3,0\quad{\rm or}\quad 2,1\quad{\rm or}\quad 1,2\quad{\rm or}\quad 0,3$$ requests on the first and the second day.

If we have on average $1.2=:\lambda$ requests per day then the probability of having $j\geq0$ requests on a given day is $$p_j={\lambda^j\over j!}e^{-\lambda}\ ,$$ and the probability of having $j$ and $k$ requests on two consecutive days is $$p_{j,k}={\lambda^{j+k}\over j!k!}e^{-2\lambda}\ .$$ It follows that $$p_{3,0}=p_{0,3}={\lambda^3\over3!0!}e^{-2\lambda},\qquad p_{2,1}=p_{1,2}={\lambda^3\over2!1!}e^{-2\lambda}\ .$$ It follows that $p_{3,0}+p_{0,3}={1\over3}(p_{2,1}+p_{1,2})$. This implies that with probability ${3\over4}$ we will observe $2,1$ or $1,2$, and with probability ${1\over4}$ we will observe $3,0$ or $0,3$. In the first case there will be three hires, in the second case just two. The expected return therefore is $$E={3\over4}\cdot 3\cdot 600+{1\over4}\cdot2\cdot 600=1650\ {\rm RM}\ ,\tag{1}$$ independently of $\lambda$.

Without knowing about the Poisson distribution we can argue as follows: Three independent random requests were received in the two days. The probability that all three of them came in on the same day is $2\cdot{1\over2^3}={1\over4}$. This at once leads to $(1)$.