A coin is tossed 5 times. What is the probability of getting at least 3 heads.

Explanation

Let `X ` denote the number of heads in `5` tosses.

`X` follows a binomial distribution with `{n}=5 ; p= `probability of getting a head `=frac{1}{2} and q=1-p=frac{1}{2}`
` P(X=r)=^{5} C_{r}(frac{1}{2})^{r}(frac{1}{2})^{5-r}, r=0,1,2 ldots 5`

The required probability `=P (`getting at least `3` heads )

`=P(X geq 3) `

`=P(X=3)+P(X=4)+P(X=5) `

`=^{5} C_{3}(frac{1}{2})^{3}(frac{1}{2})^{5-3}+^{5} C_{4}(frac{1}{2})^{1}(frac{1}{2})^{5-1}+^{5} C_{5}(frac{1}{2})^{5}(frac{1}{2})^{5-0} `

`=10(frac{1}{2})^{5}+5(frac{1}{2})^{5}+1(frac{1}{2})^{5} `

`=(frac{1}{2})^{5}(10+5+1)`

...