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The number of particles $N$ emitted by a radioactive source per a unit of time has a Poisson distribution with parameter $\lambda$. Next to the source there placed a Geiger counter that reveals every particle with probability $p$, independently. Let $M$ be the number of particles revealed by the counter. If we know that $n$ particles were emitted, find the probability that $k$ of them were revealed, that is to say, find: $p(M=k|N=n)$.

The answers only say: Bin$(k,n)$. I stated: ${p(A\cap B)\over p(B)}={{e^{-\lambda}{\lambda}^{n}\over n!}\cdot {n\choose k}p^k{(1-p)^{n-k}}\over {e^{-\lambda}{\lambda}^{n}\over n!}}=$Bin$(k,n)$, but wouldn't that mean $p(A|B)=P(A)$? Here $A=(M=k)$ and it cannot happen without using $N$. There is something here I miscomprehend, and I would appreciate it if you gave me an explanatory perspective on it.

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  • $\begingroup$ Is it right, that $P(A|B)=P(A)$. Therefore $P(A\cap B)=P(A)\cdot P(B)$. $\endgroup$ Commented Jul 20, 2015 at 13:05

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$P(A)$ is stated incorrectly. For a fixed value of $k$ you need to sum over the appropriate values of $n$.

Addition: Here are the relevant equations:

$$p(A|B)=p(M=k|N=n)={n\choose k}p^{k} (1-p)^{n-k}$$ $$p(B)=p(N=n)=e^{-\lambda} \lambda^{n}/n!$$ $$p(A)=p(M=k)=\sum_{k=n}^{\infty} p(M=k|N=n)p(N=n)=e^{-p\lambda} (p\lambda)^{k}/k!$$

So the marginal distribution of $M$ is a Poisson with mean $p\lambda$.

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  • $\begingroup$ You are correct. Is the overall computation to find the conditional probability correct, though? $\endgroup$
    – Meitar
    Commented Jul 20, 2015 at 12:53

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