They're Not All Equally Likely?

Relevant wiki: Binomial Distribution

If we tose that coin $n$ times, there will be $2^n$ possible combinations of heads and tails. If we want to find in how many of those combinations there are $m$ heads, that just will be $\binom{n}{m}$ . So, the probability of had obtained $m$ heads is $\dfrac{\binom{n}{m}}{2^n}$ . To maximize that when $n$ is even, we must have $m=\frac{n}{2}$ . When $n$ is odd, we can have $m=\frac{n-1}{2}$ or $m=\frac{n+1}{2}$ .

In this case $n=4$ , so the most likely event that happened is that we got 2 heads, with a maximized probability of $\frac{6}{16}$ .

MCQ trick: Same is valid for tails as heads. If answer is 0 head, then 4 head will also be because it means 0 tails. If answer is 1 head, 3 heads also because it is one tail. The only option is 2 heads. No contradiction for it. This trick is valid because question is single correct.