Probability: Choosing a girl from a group

[Xyius]

Homework Statement

You walk into your class the first day of classes, and you notice that
there are 30 men and 20 women in the class already. Let's suppose you decide to choose
two people from the class to be your study partners.

If you choose your study partners at random, and given that at least one of your
study partners is a woman, what is the probability of the event E that both of them
will be women?
A. 0.3167
B. 1.9%
C. 0.2405
D. 0.1901

Homework Equations

In my Solution

The Attempt at a Solution

This seems like a simple problem but I cannot seem to get the numbers available as choices.
My logic is is, W represents the event that you have picked a woman, and E represents that both of your partners will be women then.
P(E|W)=\frac{P(E \cap W)}{P(W)}
So the numerator can simplify to..
P(E|W)=\frac{P(E)}{P(W)}
This is because if E occurs, then W must have occured.
So..
P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}
and
P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}

But this doesn't work because the ratio of these two (From the formula) gives a number larger than one. Where am I going wrong? Do I use Bayes theorem?

 

[Dick]

If P(W) is the probability of choosing at least one woman, there are two ways to do that, you could choose 1 woman and 1 man, or 2 women.

 

[Xyius]

Ohh! So it would be..
P(W)=P(W|W)P(W)+P(W|M)P(M)
??
I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.

 

[Dick]

Ohh! So it would be..
P(W)=P(W|W)P(W)+P(W|M)P(M)
??
I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.

I wouldn't write it that way. Just count out the cases using combinations like you are already doing. How may ways to do each?

 

[Xyius]

So choosing 1 woman and 1 man would be
\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}
And choosing 2 women would be..

\frac{\binom{20}{2}}{\binom{50}{2}}

Plugging this in gives me 0.2405, or answer C! Thanks! :D

EDIT: Assuming that's the correct answer...

 

[Dick]

So choosing 1 woman and 1 man would be
\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}
And choosing 2 women would be..

\frac{\binom{20}{2}}{\binom{50}{2}}

Plugging this in gives me 0.2405, or answer C! Thanks! :D

EDIT: Assuming that's the correct answer...

Well, that's what I get.

 

[Ray Vickson]

Homework Statement

You walk into your class the first day of classes, and you notice that
there are 30 men and 20 women in the class already. Let's suppose you decide to choose
two people from the class to be your study partners.

If you choose your study partners at random, and given that at least one of your
study partners is a woman, what is the probability of the event E that both of them
will be women?
A. 0.3167
B. 1.9%
C. 0.2405
D. 0.1901

Homework Equations

In my Solution

The Attempt at a Solution

This seems like a simple problem but I cannot seem to get the numbers available as choices.
My logic is is, W represents the event that you have picked a woman, and E represents that both of your partners will be women then.
P(E|W)=\frac{P(E \cap W)}{P(W)}
So the numerator can simplify to..
P(E|W)=\frac{P(E)}{P(W)}
This is because if E occurs, then W must have occured.
So..
P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}
and
P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}

But this doesn't work because the ratio of these two (From the formula) gives a number larger than one. Where am I going wrong? Do I use Bayes theorem?

I W = number of women you choose, you have been asked to find the conditional probability P\{W=2|W\geq 1\}. We have P\{W=2|W\geq 1\} = \frac{P\{W = 2 \cap W \geq 1 \}}{P\{W \geq 1\}} = \frac{P\{W = 2\}}{1-P\{W=0\}}. The numerator and denominator are easily comutable using the hypergeometric distribution. The numerator is {20 \choose 2}/{50 \choose 2}, while the denominator is 1 - {30 \choose 2}/{50 \choose 2}.

RGV