Poisson Conditional Probability - Messages Received During Exams I came across a probability question.
"A student receives 6 WhatsApp messages per hour in average.  In a two-hour exam, if the student received 8 messages, what is the probability that the student received 2 messages in the first 30 minutes?".
Number of messages received per hour, k, follows a Poisson Distribution of Poi( λ = 6 messages per hour)
Here is my calculation:
$P_B$ = Prob (8 messages in 2 hours )= Poi (λ = 12 messages per 2 hours, k =8)
$$P_B = \frac{\lambda ^{k}e^{-\lambda }}{k!} = \frac{12^{8}e^{-12}}{8!}$$
$P_{A1}$ =  Prob (2 messages in 0.5 hours) = Poi ($λ_{A1}$ = 3 messages per 0.5 hours, k =2)
$$P_{A1} = \frac{\lambda_{A1} ^{k}e^{-\lambda_{A1} }}{k!} = \frac{3^{2}e^{-3}}{2!}$$
$P_{A2}$ =  Prob (8-2=6 messages in the last 1.5 hours) = Poi ($λ_{A2}$ = 9 messages per 1.5 hours, k =6)
$$P_{A2} = \frac{\lambda_{A2} ^{k}e^{-\lambda_{A2} }}{k!} = \frac{9^{6}e^{-9}}{6!}$$
Thus the absolute probability of the event E of "the student received 2 messages in first 30 minutes and then 6 messages in the next 90 mins" is：
$$ P_{abs}(E) = P_{A1}*P_{A2} $$
And conditional probability, of the event "E", given that condition of "B" happened that she/he received 8 messages in 2 hours, i.e. 120 minutes, should be:
$$ P(E|B) = \frac{P(A1\bigcap A2)}{P(B)} = \frac{P_{A1}*P_{A2}}{P(B)} $$
However, given Event B (8 message in 2 hours) and also given Event A1 (2 messages in first 30 minutes), it implies already Event A2 ( 8-2 = 6 messages, in the last 90 minutes).  Thus $P_{A2}$ is already implied by $P_B$ and $P_{A1}$.
Then one can argue that, given B, A1 and A2 are the same event, the actually conditional probability be:
$$ P(E|B) = \frac{P(A1\bigcap A2)}{P(B)} = \frac{P_{A1}}{P(B)} $$
or maybe, event A1 already implied by (Event B AND Event A2) :
$$ P(E|B) = \frac{P(A1\bigcap A2)}{P(B)} = \frac{P_{A2}}{P(B)} $$
Or we can think it this way:   Prob (2 messages in first 30 minutes, given that total 8 messages in 2 hours) = Prob ( 6 messages in the last 90 minutes of the exam),  then the answer is simply $P_{A2}$.  Right?
Anyone can help me find the right answer to this question?
Regards
$Alex$
 A: Your denominator is fine.
The numerator is $P(E \cap B)$. You have noted that $E \cap B = A_1 \cap A_2$, so $P(E \cap B) = P(A_1 \cap A_2) = P(A_1) P(A_2)$. That is all.
You cannot say that $P(A_1 \cap A_2) = P(A_1)$. I think you are perhaps thinking of $P(A_1 \cap A_2 \cap B) = P(A_1 \cap B)$, but this is something entirely different.
A: Here's another approach that illustrates a technique you may find useful when solving other problems on Poisson Processes. This approach relies on the following key fact:
The distribution of the arrival time of the $j^{\text{th}}$ WhatsApp message conditioned on the event that $8$ messages have been received is equivalent to the distribution $j^{\text{th}}$ order statistic of $8$ iid $\mathcal{U}[0,2]$ random variables.
In other words, you're asked to compute $\mathbb{P}(X_{(2)}\leq 0.5,X_{(3)}>0.5)$ where $X_1,...,X_8$ are iid $\mathcal{U}[0,2]$. Calculating such a probability requires the determination of the pdf for the bivariate random variable $\big(X_{(2)},X_{(3)}\big)$ which we'll call $f$.
To do this, first notice the distribution of $X_{(3)}$ given $X_{(2)}=x\in[0,2]$ is exactly distribution of the minimum of $6$ iid $\mathcal{U}[x,2]$ random variables whose pdf is readily calculated to be $$f_{X_{(3)}|X_{(2)}}(y|x)=\frac{6(2-y)^5}{(2-x)^6} \cdot 1_{[x,2]}$$
Next, for each $j=1,...,8$ define $I_j=1$ iff $X_j\leq x$ and $I_j=0$ otherwise. Put $I=I_1+ \dots + I_8$. Then $I\sim \text{Binomial}\Big(8,\frac{x}{2}\Big)$ and $\{X_{(2)}>x\}=\{I=0,1\}$. We get $$\begin{eqnarray*}\mathbb{P}(X_{(2)}\leq x) &=& 1- \mathbb{P}(X_{(2)}>x) \\ &=& 1- \mathbb{P}(I=0)-\mathbb{P}(I=1) \\ &=& 1- \Big(1-\frac{x}{2}\Big)^8 - 8 \Big(\frac{x}{2}\Big) \Big(1-\frac{x}{2}\Big)^7 \\ &=& \Big(\frac{x}{2}-1\Big)^7 \Big(\frac{7x}{2}+1\Big)\end{eqnarray*}$$ Taking a derivative yields the pdf of $X_{(2)}$. $$f_{X_{(2)}}(x)=\frac{7}{32}x(2-x)^6 \cdot 1_{[0,2]}$$ So the joint pdf of $\Big(X_{(2)},X_{(3)}\Big)$ (which we originally called $f$) is $$f(x,y)=f_{X_{(3)}|X_{(2)}}(y|x)f_{X_{(2)}}(x)=\frac{21}{16}x(2-y)^5\cdot 1_{S}$$ Finally, $$\mathbb{P}(X_{(2)}\leq 0.5,X_{(3)}>0.5)=\int_{1/2}^2 \int_0^{1/2}f(x,y)\mathrm{d}x\mathrm{d}y=\frac{5103}{16384}$$
