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's dominant hand might be either right or left. It is assumed that 90% of the population is right-handed.

Therefore, `p=frac{90}{100}=frac{9}{10}`

and `q=1-p=1-frac{9}{10}=frac{1}{10}, n=10`

Therefore, X has a binomial distribution with `n=10,p=frac{9}{10} and q=frac{1}{10}`

Then, `P(X = x) = ^nC_x q^n - xp^x, x = 0, 1, 2, ...n`

`P(X = r) = ^10C_r(9/10)r(1/10)^(10 - r`)

Hence,

`P(X<=6)=1-sum_(r=7)^10 ^10C_r(9/10)r(1/10)^(10 - r)`