Question

A city council consists of six Democrats and six Republicans . If a committee of six people is selected , find the probability of selecting two Democrats and four Republicans

Answer 1

There are 6 democrats and 6 republicans.

The possible ways of selecting 2 democrats is given by the combination:

$_6C_2=\frac{6!}{2!(6-2)!}=15$

The possible ways of selecting 4 republicans is:

$_6C_4=\frac{6!}{4!(6-4)!}=15$

Now, using the fundamental counting principl, the number of ways of selecting 2 democrats and 4 republicans is:

$_6C_2\cdot_6C_4=15\cdot15=225$

The total number of possible combinations is given by:

$_{12}C_6=\frac{12!}{6!(12-6)!}=924$

Finally the probability is given by:

$P=\frac{\text{ number of ways of selecting 2 democrats and 4 republicans}}{\text{ Total number of possible combinations}}$

Therefore:

$P=\frac{225}{924}=\frac{75}{308}$

Hence the probability is 75/308