1

Customers arrive at a facility according to a Poisson process $N(t)$ of rate $\lambda = 5.5$ customers/hour. Each customer is admitted to the facility with probability $p=0.6$. All customers, who are not admitted, leave and do not come back. Let $X(t)$ be the number of customers admitted from time $0$ to time $t$. Compute the following:
$(a)$ The mean value $\mathbb E[X(10)]$
$(b)$ The probability $\mathbb P(X(10) = 31)$
$(c)$ The conditional probability $\mathbb P(X(10) = 31 \mid N(10) = 54)$.

So far I have figured out that $\mathbb E[X(10)] = p\lambda t = 33$.

My attempt for $(b)$ is $$\mathbb P(X(10)=31) = 0.6\cdot\frac{(5.5\cdot10)^{31}e^{-5.5\cdot10}}{31!},$$ which is wrong. I hope someone will help me figure out the next parts and lead me in the right direction

calculatormathematical
  • 2,222
waterr
  • 67

1 Answers1

1

The idea is that, since people are arriving according to a Poisson process, and each person is admitted with the same independent probability, people are also being admitted according to a Poisson process.

Since an average of $5.5$ people are arriving each hour, and an average of $60\%$ of them are admitted, people are being admitted at an average rate of $0.6 \cdot 5.5 ~\text{people}/\text{hr} = 3.3 ~\text{people}/\text{hr}$.

Thus, $X(10)$, which is the number of admissions between time $0$ and $10$ hours, is Poisson-distributed with parameter $10 \text{hr} \cdot 3.3 /\text{hr} = 33$. We conclude that

$$\mathbb{P}(X(10) = 31) = \boxed{e^{-33} \frac{33^{31}}{31!}}.$$

This probability is approximately $0.0672$.

diracdeltafunk
  • 14,911
  • For c), should I assume that $X(t)$ and $N(t)$ are independent of each other and so $P(X(10) = 31 \cap N(10) = 51) = P(X(10)=31) ⋅ P(N(10) = 51)$ and so the result would just be $P(X(10) = 31)$? – waterr Mar 04 '24 at 00:59
  • No, $X(t)$ and $N(t)$ are absolutely not independent. For example, it is possible that $X(1)=1$ and it is possible that $N(1)=0$, but it is not possible to have both $X(1)=1$ and $N(1)=0$. – diracdeltafunk Mar 04 '24 at 01:25
  • So since 31 customers are admitted out of the 54 that arrived, with a success being admitted and failure being not admitted, that would mean that the joint probability would be $Binomial(31;51,0.6)$? – waterr Mar 04 '24 at 02:11
  • Yes, that's right! – diracdeltafunk Mar 04 '24 at 03:38