Probability of hitting the target for the nth time on the mth throw

Question

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.

Answer 1

Presumably each throw is independent of one another (something which is critical to note in order to continue the problem because without this assumption there is not enough information to answer).

Now... the event we are interested in calculating the probability of as I understand your question to be is that among the first nine throws, exactly four of the throws were hits and also the tenth throw is a hit.

The results of the first nine throws is going to be independent of the result of the tenth throw, so we can find the probabilities of each and multiply the results. To calculate each, we can use a binomial distribution.

This gives:

$$\binom{9}{4}0.5^4\times 0.5^{9-4} ~~~\times ~~~0.5$$

Or simplified:

$$\binom{9}{4}\times 0.5^{10}=\dfrac{63}{512}$$

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If your question is instead, what is the probability that within ten throws, at least five of the throws are hits, then this will be

$$\sum\limits_{k=5}^{10}\binom{10}{k}0.5^k\times 0.5^{10-k}$$

or simplified

$$\dfrac{319}{512}\approx 0.623$$

It is equally likely that there are strictly more hits than misses versus having strictly more misses than hits here. It is also possible that there are an equal number of hits and misses, and so the probability that we have at least as many hits as misses (i.e. 5 or more) will be strictly more than $0.5$.

Compare this to if there are an odd number of throws, where it is no longer possible to have an equal number of hits and misses. If we throw nine times, we find that since it is just as likely to hit as to miss, then the probability to get more hits than misses will be $0.5$ without needing to run the calculations manually.

Answer 2

In $10$th throw, the probability doesn't change. So his chances of hitting the target would still be $1/2$. It doesn't matter how many times he has hit the target before. Even if he misses the target a million times in a row, the probability at million+1 throw would still be $1/2$. Lookup gambler's fallacy.

Answer 3

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

$$ \binom{9}{4}\dfrac{1}{2^{10}} = \dfrac{63}{512} \approx 12.3\% $$

Answer 4

Probability of hitting target is $~\frac{1}{2}~$ and therefore probability of hitting target is $~1- \frac{1}{2}=\frac{1}{2}~$.

So $~P~=~$$\{$ probability of hitting target $~4~$times in $~9~$ throw $\}~-~\{$ probability of hitting target in $~10^{th}~$ throw $~\}$

$~~~~~~~~~~=~\left\{^9C_4 ~\times~ \left(\frac{1}{2}\right)^4~\times~ \left(\frac{1}{2}\right)^5\right\}~\times~\left(\frac{1}{2}\right)$

$~~~~~~~~~~=~^9C_4 ~\times~ \left(\frac{1}{2}\right)^{10}$

$~~~~~~~~~~=\frac{63}{2^9}$