Question

A focus group of 12 people is to be chosen randomly from among 24 right-handed people and 5 left-handed people. In order to find the probability that 3 of the people chosen are right-handed, you should use

Answer 1

Let 1 and 2 mean that a person is right-handed or left-handed, respectively.

Consider the probability space as the different groups of 12 people that can be formed with the 29 total people. {(1,1,1,1,1,...1),(2,1,1,1,1,...,1),...}

We need to use the binomial distribution in order to find the answer.

Consider X to be the number of right-handed people in the group.

The probability of X=3 is then:

`Pr(X=3)=\text{ binomial coefficien(}12,3\text{)}((24)/(29))^3(1-(24)/(29))^(12-3)`

We used the formula

`\begin{gathered} Pr(X=k)=\text{ binomial coefficient(n,k)}\cdot p^k(1-p)^(n-k) \\ \text{where } \\ k=3 \\ n=12 \\ p=(24)/(29) \end{gathered}`

Finally, we need to simplify the expression, as shown below

`\Rightarrow P(X=3)=220((24)/(29))^3((5)/(29))^9=0.0000167`

This is the answer one obtains using the binomial distribution; nevertheless, the actual probability is equal to zero because it is not possible to form a group of 12 people that only contains 3 right-handed people as there are only 5 left-handed people (3+5=8).