In tossing 10 coins the probability of getting exactly 5 heads

For one toss, the probability of getting one head is 1/2.

For two tosses, the probability of getting two heads is (1/2)(1/2) = 1/4
(that is you need to have one head, then one head)

For two tosses, the probability of getting one head is (1/2)(1/2) + (1/2)(1/2) = 1/2
(that is, you need to have either one first head, then one tail, or one first tail, then one head)

Can you try for your problem now? Post the answer you think is correct. :)

Work it through like this: first, imagine you wanted to get 5 heads in a row followed by 5 tails in a row. The probability of that happening is (1/2)^10. Do you see why?

But to win this game you don't need this precise order of tosses. You succeed no matter what order of heads and tails there is. So you multiply this by the number of ways 5 items out of 10 can be arrannged - because we have 10 items (10 tosses), and we can move the 5 heads around to any of those ten toss positions and still "win." So the final answer is:

C(10,5) x (1/2)^5

In general, given probability p for something to occur (like getting heads), then the probability of getting k successes out of n trials is:


P(\text{k successes}) = C(n,k)p^k (1-p)^{n-k}


where C(n,k) is:


C(n,k) = \frac {n!} {k!(n-k)!}

Note - to avoid any confusion with notation here - the notation of the combination of k things out of n may also be written as follows:


C(n,k)\ =\ _n C _k \ =\ \left( n \\ k \right)