Question

Three people are selected for a committee from 4 Democrats and 3 Republicans. What is the probability the committee contains both Democrats and Republicans given that it contains at least two Democrats?

Answer 1

Out of $4$ Democrats and $3$ Republicans a committee of $3$ people can be select in$^7C_3$ ways.

Therefore sample space $S$ contain $^7C_3$ number of elements.

Let $A$ be an event such that at least two Democrats choose where committee contains both Democrats and Republicans.

Now out of $3$ person at least $2$ Democrats can be chosen such a way

(1) 2 Democrats $1$ Republicans

(2) 3 Democrats

Therefore $2$ Democrats and $1$ Republicans can be choose in $(^4C_2×{^3C_1})$ ways.

And $3$ Democrats can be choose in $^4C_3$ ways.

So number of elements in favour of $A$ is $[(^4C_2×{^3C_1})+^4C_3]=22$ .

$\therefore P(A)=\frac{22}{^7C_3}=\frac{22}{35}$

Let $B$ be an event such that committee of both Democrats and Republicans.

This can be chosen in such a way

(1) $1$ Democrat $2$ Republicans

(2) $2$ Democrats $1$ Republican

First can be choose in $(^4C_1×{^3C_2})$ ways and second can be choose in $(^4C_2×{^3C_1})$ ways.

We have to find $P(B/A)$ .

Now $P(B/A)=\frac{P(B\cap A)}{P(A)}$

Here number of elements in favour of $(B\cap A)$ is $(^4C_2×{^3C_1})=18$ .

$\therefore P(B\cap A)=\frac{18}{^7C_3}=\frac{18}{35}$ .

Hence $P(B/A)=\frac{P(B\cap A)}{P(B)}=\frac{\frac{18}{35}}{\frac{22}{35}}=\frac{18}{22}=\frac{9}{11}$.

Which is the required probability.