Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Explanation

Let `X` be the number of people that are -handed in the sample of `10` people.

`X` follows a binomial distribution with `n=10,`

`p=90 `

`P(X=r)=^{10} C_{r}(0.9)^{r}(0.1)^{10-r}`

Probability that at most ` 6` are - handed` =P(X leq 6)`

`=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6) `

`=1-P(X=7)+P(X=8)+P(X=9)+P(X=10) `

`=1-sum_{r=7}^{10}{ }^{10} C_{r}(0.9)^{r}(0.1)^{10-r}`