Probability of hitting the target for the nth time on the mth throw A boy is throwing stones at a target. Probability of him hitting the target is $\frac{1}{2}$ . Then the probability of him hitting the target for the  5th time in 10 throws is ?
The answer given is 1/2 .

Now ,the reason given is that no matter how many throws he takes or no matter how many times he hits (be it 10 times in 10 throws or 6 times in 7 throws or all initial 5 throws hitting the target out of 10 throws ) since the probability of him hitting the target is always 1/2 therefore the net probability will always be 1/2.
But while putting it into the binomial distribution formula the answer coming is different.

In probability theory this is called a negative binomial distribution. What it does is answer the question of what is the probability that you will have $n$ successes after $m$ attempts. Technically this notation is wrong, but I modified it to match your question. 
For example, you could say that in the event of a Best-of-$3$ rock paper scissors game, what are the odds that the match will be over after only $2$ rounds? In your case the question is, what are the odds that he will hit the target for the $n$th time on the $m$th throw. The formula is as follows
$$
\binom{m - 1}{n - 1}p^n(1-p)^{m-n}
$$
The $\binom{m-1}{n-1}$ comes from the fact you need to choose the location of all successes, and already know that the final throw has to be a hit to make sure your condition is satisfied. So in your example with $p=\frac{1}{2}$, the formula for any $m$ and $n$ are
$$
\binom{m - 1}{n - 1}\dfrac{1}{2^m}
$$
so for $n=5,m=10$ we find
$$
\binom{9}{4}\dfrac{1}{2^{10}} = \dfrac{63}{512} \approx 12.3\%
$$
