Eight coin are thrown simultaneously. Find the chance of obtaining at least six heads.

Explanation

Let `X` be the number of heads in tossing `8` coins.

`X` follows a binomial distribution with `n=8;`

`p=frac{1}{2} and q=frac{1}{2}`

`P(X=r)=^{8} C_{r}(frac{1}{2})^{r}(frac{1}{2})^{8-r}=^{8} C_{r}(frac{1}{2})^{8}`

Probability of obtaining at least `6` heads `=P(X geq 6)`

`=P(X=6)+P(X=7)+P(X=8)`

`=^{8} C_{6}(frac{1}{2})^{8}+^{8} C_{7}(frac{1}{2})^{8}+^{8} C_{8}(frac{1}{2})^{8}`

`=frac{1}{2^{8}}(28+8+1)`

...