So this question wasn't answered before:

1.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.

Thanks!!!!!!!!!!

MIRB16 Feb 27, 2018

#1**+1 **

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 .

ElectricPavlov Feb 27, 2018

edited by
Guest
Feb 27, 2018

#2**0 **

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.

Alan Feb 27, 2018

#3**0 **

**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. **

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

heureka Feb 27, 2018