Probability of $k$ successes in $n$ trials.

Question

If you have a $n = 10$ trials and you want to find the probability for $k = 2$ successes you'd use $$P(x=k)={n\choose k}(p)^k(1-p)^{n-k}.$$ If we take $p = 1/5$ we get $0.3019...$. If we take $p = 1/4$ we get $0.2815...$. Why is the chance of getting 2 successes in 10 trials higher if the probability is lower? This seems counterintuitive so I assume I've done something wrong?

Answer 1

Here is an illustration of the binomial probability distributions with $p=\frac15$ in orange and $p=\frac14$ in blue, with $n=10$ for both.

You can see that the blue probabilities are greater than the orange probabilities for higher values of the random variable as you might intuitively expect, and so the opposite for lower values. They have to cross somewhere and in this case it is between $2$ and $3$.

For a binomial distribution, the probability for seeing $x$ successes out of $n$ is maximised when $p=\frac xn$, so $\frac xn$ is the maximum likelihood estimator for $p$. In this case with $n=10$ and $x=2$, the probability is maximised when $p=\frac2{10}=\frac15$.