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=(6!)/(2!(6-2)!)=15`

The possible ways of selecting 4 republicans is:

`_6C_4=(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=(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=(225)/(924)=(75)/(308)`

Hence the probability is 75/308