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"Customers arrive at a bank according to a Poisson process with rate 6 per hour.

State (together with a proof) clearly the (conditional) probability mass function of the numbers of customers arrived during the first 20 minutes, given that 10 customers have arrived during the first hour."

I did this:

$$ P(X_{\frac{1}{3}} = x | X_1 = 10) $$ From the definition of joint probability, we know

$$ P(X_{\frac{1}{3}} = x , X_1 = 10) = P(X_{\frac{1}{3}} = x | X_1 = 10) P(X_1 = 10)$$

Re-aranging gives:

$$ P(X_{\frac{1}{3}} = x | X_1 = 10) = \frac{ P(X_{\frac{1}{3}} = x , X_1 = 10)} {P(X_1 = 10)}$$

Let's first calculate $P(X_{\frac{1}{3}} = x , X_1 = 10)$:

This becomes:

$$ P(X_{\frac{1}{3}} = x , X_{\frac{2}{3}}= y) = P(X_{\frac{1}{3}} = x) P(X_{\frac{2}{3}} = y) \hspace{1cm} y = 10 - x $$

Using the Poisson formula, you get:

$$P(X_{\frac{1}{3}} = x) = \frac{e^{-2}2^x}{x!} \hspace 2cm P(X_{\frac{2}{3}}= y) = \frac{e^{-4}4^y}{y!}$$

And so

$$ P(X_{\frac{1}{3}} = x , X_{\frac{2}{3}}= y) = e^{-6} \frac{2^x4^y}{x!y!} $$

Also, using the Poisson formula, we get:

$$ P(X_1 = 10) = \frac{e^{-6}6^{10}}{10!} $$

So

$$\frac{ P(X_{\frac{1}{3}} = x , X_1 = 10)} {P(X_1 = 10)} = \frac{e^{-6} \frac{2^x4^y}{x!y!}}{\frac{e^{-6}6^{10}}{10!}} $$

Which I get to be

$$ \frac{10! 2^x 4^y}{x!y! 6^{10}}$$

But in the answers, it says it should be

$$ \binom{10}{x} \left( \frac{1}{3} \right)^x \left( \frac{2}{3} \right)^{10 - x} $$

Why is this?

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I assume in your question you mean $6^{10}$ as opposed to $10^6$. So, let's start with what you have: $$\frac{10! 2^x 4^{10-x}}{x!(10-x)!6^{10}} = \frac{10!}{x!(10-x)!}\times\frac{2^x 4^{10-x}}{6^{10}}.$$ The first term is $\binom{10}{x}.$ Let's pull $2^{10}$ out of the top and bottom of the second term and cancel them, leaving us with $$\binom{10}{x} \frac{1^x 2^{10-x}}{3^{10}} = \binom{10}{x} \frac{1^x}{3^x} \frac{2^{10-x}}{3^{10-x}} = \binom{10}{x}\left(\frac13\right)^x\left(\frac23\right)^{10-x}$$ which is the answer you want.

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  • $\begingroup$ If you have $2^x$ and you divide this by $2^{10}$, does that not become $1^{x - 10}$? $\endgroup$ – Kaish Dec 19 '12 at 15:14
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    $\begingroup$ No, it would become $2^{x-10}$. But I also pulled $2^{10-x}$ out of the $4^{10-x}$, so in total I pulled $2^{10}$ out of the top. Remember $1^{\text{whatever}} = 1$. $\endgroup$ – Jonathan Christensen Dec 19 '12 at 15:34

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