Question

According to data from the American Medical Association, 10% of us are left handed. If three people are randomly selected, find the probability that they are all left-handed. What probability distribution can we use to answer this question

Answer 1

For each person, there are only two possible outcomes. Either they are left handed, or they are not. The probability of a person being left-handed is independent from other people. So we use the binomial probability distribution to solve this question.

0.1% probability that they are all left-handed.

Explanation:

For each person, there are only two possible outcomes. Either they are left handed, or they are not. The probability of a person being left-handed is independent from other people. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

10% of us are left handed.

This means that
`p = 0.1`

If three people are randomly selected, find the probability that they are all left-handed.

This is P(X = 3) when n = 3. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(3,3).(0.1)^(3).(0.9)^(0) = 0.001`

0.1% probability that they are all left-handed.