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:
[tex]Pr(X=3)=\text{ binomial coefficien(}12,3\text{)}(\frac{24}{29})^3(1-\frac{24}{29})^{12-3}[/tex]We used the formula
[tex]\begin{gathered} Pr(X=k)=\text{ binomial coefficient(n,k)}\cdot p^k(1-p)^{n-k} \\ \text{where } \\ k=3 \\ n=12 \\ p=\frac{24}{29} \end{gathered}[/tex]Finally, we need to simplify the expression, as shown below
[tex]\Rightarrow P(X=3)=220(\frac{24}{29})^3(\frac{5}{29})^9=0.0000167[/tex]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).
