if 10 coins are tossed what is the probability of getting 7 heads?

For this, you have to use the binomial distribution.

Let X be the event that a head is obtained.

X ~ B (10, 0.5)

10 because there are 10 tosses and 0.5 because the probability of getting a head in each toss is 0.5.

The, find P(X = 6).

Can you work this out?

EDIT: I made a typo. This should read as:
Then, find P(X = 7).

You can think this problem through as follows:

You toss a coin 10 times and want to get exactly 7 heads and 7 tails. The probability of getting either with each flip is 0.5. So to get 7 heads and then 7 tails in that order is:


( \frac 1 2 ) ^ 7 ( \frac 1 2 ) ^3 = ( \frac 1 2 ) ^{10}


But in this problem we don't care about order. So multiply this by the number of ways that you can arrange the 7 heads and 3 tails in 10 coin flips - that number is C(10,3), or:

\frac {10!} {(10-3)! 3!} = \frac { 10 \cdot 9 \cdot 8 } {3 \cdot 2 \cdot 1}


So the probability of getting 7 heads and 3 tails in any order is:


C(10,3) (\frac 1 2 )^{10}