Hi!

I seem to almost never get these correct.....still trying to get this probability stuff to 'click' in my mind. Here is my best guess:

Ways to fill the committee   5 c 3 = 10  possible ways to pick from the pool of 5 people

    of these 10 selections there are 3 possibilities that have 2 women and one man

          so....so far  3/10 chance for 2 women and one man on the committee

With any of these choices, there is a 2/3 chance of picking a woman for the chair position

so  3/10 x 2/3 =  6/30  =1/5  chance  for    2 women 1 man   and woman chair .

Here’s another way of looking at it:

Suppose a, b and c are men and x and y are women. The total possible arrangements are:

abc

abx 

aby 

acx 

acy 

axy 

bcx 

bcy 

bxy 

cxy

Of these ten, three have two women and one man, so Probability of this style committee is 3/10.

Probability of such a committee having a woman as chair is 2/3.

Hence overall Probability is (3/10)*(2/3) = 1/5 as EP obtained.

A committee of three people is to be randomly selected from a group of three men and two women, and the chairperson will be randomly selected from the committee. What is the probability that the committee will have exactly two women and one man, and that the chairperson will be a woman?

Express your answer as a common fraction. 

\(\begin{array}{|rcll|} \hline && \left[\dfrac{ \binom22\binom31 } {\binom53} \right]\times \left[ \dfrac{ \binom21\binom10 } {\binom31} \right] \\ &=& \left[ \dfrac{ 1\cdot 3 } {\binom52} \right]\times \left[ \dfrac{ 2\cdot 1 } {3} \right] \quad & | \quad {\binom53} = {\binom52} = \dfrac {5}{2}\cdot \dfrac{4}{1} = 10 \\ &=& \left[ \dfrac{ 3 } {10} \right]\times \left[ \dfrac{ 2 } {3} \right] \\ &=& \dfrac{ 2 } {10} \\ &=& \dfrac{ 1 } {5} \\ \hline \end{array} \)