Probability Question using Poisson distribution On average, an employee receive 25 emails each day, of which 60% are ‘spam’. What is the probability that the employee will receive exactly 15 ‘spam’ emails tomorrow?
My methodology is such:
$$
\frac {\lambda^{15}}{15!}e^{-\lambda} = \frac{15^{15}}{15!}e^{-15} = 0.1024 = 10.24\%
$$
I'm not certain about my answer. All I know is that, I have to use Poisson distribution.
Additional content:
I used binomial distribution to figure out the probability but the answer in this one is different from the one above. Not sure which method to rely on now.
$$
P(S=15)=\left(\begin{matrix}
        25  \\
        15 
        \end{matrix}\right)(0.4)^{15} (0.6)^{10} = 0.021 = 2.1\%
$$
 A: The binomial distribution as it was used in the question is not correct unless the distribution of the number of emails received in a day is not random; i.e., exactly 25 emails are received each day, and the probability that any given email is spam is exactly 60%.
If, however, we assume that the number of emails received in a day is a random variable $N$ which is Poisson distributed with mean $\lambda = 25$, the conditional distribution of the number of spam emails received on a given day with $N = n$ total emails is $$X \mid N = n \sim \operatorname{Binomial}(n,0.6).$$  The unconditional distribution is therefore $$\Pr[X = x] = \sum_{n=x}^\infty \Pr[X = x \mid N = n]\Pr[N = n] = \sum_{n=x}^\infty \binom{n}{x} p^x (1-p)^{n-x} e^{-\lambda} \frac{\lambda^n}{n!} = e^{-p\lambda} \frac{(p \lambda)^x}{x!},$$ which is of course itself Poisson with rate parameter $p\lambda$.  So the probability that $X = 15$ is simply $$\Pr[X = 15] \approx 0.102436.$$  In comparison, if $N = 25$ and we use the binomial model, $$\Pr[X = 15] \approx 0.161158.$$  (Your calculation is not correct because you've switched $p$ and $1-p$.)
Under a different distributional assumption for the number of emails received in a day, the answer will be different.  $N$ need not be $25$, nor does it need to be Poisson.  It could be negative binomially distributed; it could be uniform; it could be any discrete distribution whose support is a subset of the nonnegative integers.  The question as it is posed does not impose such a distribution nor does it imply one--it only supposes that its mean is $25$.
A: Assuming the Poisson distribution is applicable, this is correct. However there may be some uncertainty as to whether the Poisson distribution is really applicable. Are the spam emails recieved entirely at random, and independently?
