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

A person can be either right-handed or left-hended. It is given that `90%` of the people are right-handed. Therefore,
`p=P("right-handed")=9/10`
`q=P("left-handed")=1-9/10=1/10`
Using binomial distribution, the probability that at most six people are right-handed is given by
`underset(r=0)overset(6)sum""^(10)C_(r)p^(r)q^(n-r)underset(r=0)overset(6)sum""^(10)C_(r)((9)/(10))^(r)((1)/(10))^(10-r)`
Therfore, the probability that at most 6 people are right-handed
`underset(r=7)overset(10)sum""^(10)C_(r)(0.9)^(r)(0.1)^(10-r)`