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Each second, an ounce of a radioactive substance poissonium emits 5 alpha particles on average. Approximate the probability that exactly 4 alpha particles are emitted by an ounce of poissonium over the next second.

I am told that I can use a poisson approximation for this problem, and that $\lambda = 5$:

$$ X \rightarrow \text{ The number of alpha particles emitted by an ounce of poissonium in 1 second.} \\ P(X=4) = e^{-\lambda}\frac{\lambda^4}{4!} = e^{-5}\frac{5^4}{4!} \approx 0.175 $$

I don't understand where $\lambda = 5$ came from though. If $\lambda = np$, then what is $n$, and what is $p$?

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$n$ is the number (very large) of atoms in the ounce, and $p$ is the probability a particular atom decays in the next second. –  André Nicolas Oct 8 '12 at 21:13
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The parameter for a poisson process is the average number of events per unit time. This is given in the question and doesn't need to be calculated (5 events per second). Sometimes you need $n$ and $p$ to compute $\lambda$ - but in this physical case the average rate of emission is a quantity which can be determined by experiment/observation, while the number of particles and the probability of decay per second are not so accessible. –  Mark Bennet Oct 8 '12 at 21:23
    
Although three answers have been posted, so far I'm the only one who's up-voted the question. –  Michael Hardy Oct 9 '12 at 18:39
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3 Answers

up vote 1 down vote accepted

In the formula $$ e^{-\lambda}\frac{\lambda^k}{k!} $$ $\lambda$ is the estimated rate times the time interval, that would be the expected number of occurrences during the given time interval. In this case, the average is $5$ per second and the time period is $1$ second, so $\lambda=5$.

Say we wanted to compute the probability that only $2$ events happen in the next $2$ seconds. Then $\lambda=10$ ($2$ seconds times $5$ per second) and so we get $$ e^{-10}\frac{10^2}{2!}=0.00227 $$

Explanation:

Let $\lambda$ be the expected number of events in a given time interval. Then the Poisson assumption is that the probability that an event will happen in a given $\frac1n$ slice of that time interval is $\frac\lambda n$. Let's compute the probability that $k$ events happen in that same time interval. The chance that more than one event happens in one $\frac1n$ slice of the time interval is so small that we can ignore it. The binomial estimate says that the probability that we have exactly $k$ events is $$ \binom{n}{k}\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k $$ That is, $\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k$ is the probability of $n-k$ non-events and $k$ events. Furthermore, there are $\binom{n}{k}$ different ways to arrange the $k$ events and $n-k$ non-events.

Now, let $n\to\infty$: $$ \begin{align} &\lim_{n\to\infty}\binom{n}{k}\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k\\ &=\lim_{n\to\infty}\frac nn\frac{n-1}n\frac{n-2}n\dots\frac{n-k+1}{n} \left(1-\frac \lambda n\right)^{n-k}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\frac{\lambda^k}{k!} \end{align} $$

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@MichaelHardy: Use a zero before the decimal point for numbers less than one. For example: $$ t = 0.40 $$ However, do not use a zero before the decimal point when the number cannot be greater than one. This occurs with correlations, proportions and levels of statistical significance. For example: $$ r = .27, p < .01 $$ "Tables" in Taylor and Francis Online - Instructions for Authors and APA Style Blog –  robjohn Oct 9 '12 at 19:38
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For poisson random variable $\lambda$ is not only the parameter, but also the expected value (i.e average) which is given in your question. As pointed in comments, it might be easier to find average rate of emmission per secon than number of particles,etc

Now I suppose you were talking about Poisson approximation to binomial. It is not the case here but if you had been given that some process follows $\mathrm{Binomial}(n,p)$ with $np$ constant,you could apply that. They are different situatons.

Assume that question had been, the sample contains $1 \times 10^6$ alpha particles and probability of each particle being emmitted is $5 \times 10^{-6}$. Then you could use poisson approximation to reduce it to previous problem as this is binomial with $np=5$.

Hope that helps.

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André Nicolas suggested that $n$ is the number of atoms and $p$ is the probability that any atom decays within the next second. That's OK as far as it goes, but it doesn't take into account the infinite divisibility of time. Suppose you divide the second into a million equal intervals. If the probability of a decay in any of those intervals is $5/(1\text{ million})$ and the events of decay in different intervals are independent, then the expected number of decays is $5$ and the number of decays is a binomially distributed random variable with parameters $n=1\text{ million}$ and $p=5/(1\text{ million})$. But that's not quite infinite divisibility. Let's get a bit closer: divide each of those $1\text{ million}$ intervals into another $1\text{ million}$ smaller intervals, so there are $(1\text{ million})^2$ intervals, with probability $5/((1\text{ million})^2)$ of a decay in each interval. This of course introduces the possibility of more than one decay in an interval of length $1/(1\text{ million})$, but the probability that that happens at least once is negligible, so our million intervals was a reasonable approximation to the situation with a million-squared intervals. As the number of intervals approaches $\infty$, the binomial distribution approaches a Poisson distribution.

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$$ \lim_{n\to\infty}\binom{n}{k}\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k=e^{-\lambda}\frac{\lambda^k}{k!} $$ –  robjohn Oct 9 '12 at 3:34
    
If we want to get into that, let's write $\dbinom{n}{k}\left(1-\frac{\lambda}{n}\right)^{n-k}\left(\frac\lambda n\right)^k$ $=\underbrace{\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}}$ $\underbrace{\left(1-\frac\lambda n\right)^n}$ $\frac{\lambda^k}{k!}$ $\underbrace{\left(1-\frac\lambda n\right)^{-k}}$. Then finding the limits of the things over the underbraces is routine. Two of them approach $1$ and one of them approaches $e^{-\lambda}$. –  Michael Hardy Oct 9 '12 at 17:21
    
That is done in the Explanation to my answer :-) –  robjohn Oct 9 '12 at 17:25
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