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

Answer:

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.

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.