Question

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?

Answer 1

Let X be the number of people that are right-handed in the sample of 10 people.
X follows a binomial distribution with n = 10,

\[p = 90 % = 0 . 9 and q = 1 - p = 0 . 1\]

\[P(X = r) = ^{10}{}{C}_r (0 . 9 )^r (0 . 1 )^{10 - r} \]

\[\text{ Probability that at most 6 are right - handed } = P(X \leq 6)\]

\[ = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)\]

\[ = 1 - {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}\]

\[ = 1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}\]