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.
$$ P(S=15)=\left(\begin{matrix} 25 \\ 15 \end{matrix}\right)(0.4)^{15} (0.6)^{10} = 0.021 = 2.1\% $$
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 $25$.