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How would i work out $Pr(x=2 | x>1)$
when $E(x)=3$

would it be the answer of $x=2$ divided by answer of $x>1$ or the other way around, as it could mean the answer of $x=2$ is a divisor of the answer of $x>1$

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1 Answer 1

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Remember Bayes' Theorem:

$$\mathbb{P}(X\mid Y)=\frac{\mathbb{P}(Y\mid X)\mathbb{P}(X)}{\mathbb{P}(Y)}$$

In this case, $X$ is the event that $x=2$ and $Y$ is the event that $x>1$. It is evident that $x=2 \implies x>1$, therefore $\mathbb{P}(x>1\mid x=2)=1$, so we have:

$$\mathbb{P}(x=2\mid x>1)=\frac{\mathbb{P}(x=2)}{\mathbb{P}(x>1)}$$

Which for $x\sim \text{Po}(3)$ gives:

$$\mathbb{P}(x=2\mid x>1)=\frac{\frac{9}{2\rm{e}^{3}}}{\frac{\rm{e}^{3}-4}{\rm{e}^{3}}}=\frac{9}{2(\rm{e}^{3}-4)}\approx 0.2798$$

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