Probability: Choosing a girl from a group

In summary, the probability of choosing two women as study partners from a class of 30 men and 20 women, given that at least one of the study partners is a woman, is 0.2405. This can be calculated using the hypergeometric distribution, where the numerator is the probability of choosing 2 women out of 50 people, and the denominator is 1 minus the probability of choosing 0 women out of 50 people.
  • #1
Xyius
508
4

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.
[tex]P(E|W)=\frac{P(E \cap W)}{P(W)}[/tex]
So the numerator can simplify to..
[tex]P(E|W)=\frac{P(E)}{P(W)}[/tex]
This is because if E occurs, then W must have occured.
So..
[tex]P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]
and
[tex]P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}[/tex]

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?
 
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  • #2
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.
 
  • #3
Ohh! So it would be..
[tex]P(W)=P(W|W)P(W)+P(W|M)P(M)[/tex]
??
I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.
 
  • #4
Xyius said:
Ohh! So it would be..
[tex]P(W)=P(W|W)P(W)+P(W|M)P(M)[/tex]
??
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?
 
  • #5
So choosing 1 woman and 1 man would be
[tex]\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}[/tex]
And choosing 2 women would be..

[tex]\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]

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

EDIT: Assuming that's the correct answer...
 
  • #6
Xyius said:
So choosing 1 woman and 1 man would be
[tex]\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}[/tex]
And choosing 2 women would be..

[tex]\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]

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.
 
  • #7
Xyius said:

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.
[tex]P(E|W)=\frac{P(E \cap W)}{P(W)}[/tex]
So the numerator can simplify to..
[tex]P(E|W)=\frac{P(E)}{P(W)}[/tex]
This is because if E occurs, then W must have occured.
So..
[tex]P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]
and
[tex]P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}[/tex]

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 [itex]P\{W=2|W\geq 1\}. [/itex] We have [tex]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\}}.[/tex] The numerator and denominator are easily comutable using the hypergeometric distribution. The numerator is [itex] {20 \choose 2}/{50 \choose 2},[/itex] while the denominator is [itex] 1 - {30 \choose 2}/{50 \choose 2}.[/itex]

RGV
 

1. What is the probability of choosing a girl from a group of 10 people?

The probability of choosing a girl from a group of 10 people depends on the gender distribution of the group. If the group is evenly split between boys and girls, the probability would be 50%. However, if there are more girls than boys in the group, the probability would be higher and vice versa.

2. If there are 4 girls and 6 boys in the group, what is the probability of choosing a girl?

The probability of choosing a girl would be 4 out of 10 or 40%. This can be calculated by dividing the number of girls (4) by the total number of people in the group (10).

3. How does the number of people in the group affect the probability of choosing a girl?

The larger the group, the lower the probability of choosing a girl. This is because as the group size increases, the number of boys and girls becomes more evenly distributed, resulting in a lower probability of choosing a specific gender.

4. What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and predicts the likelihood of an event occurring. Experimental probability is based on actual outcomes from trials or experiments and can differ from theoretical probability due to chance or external factors.

5. Can the probability of choosing a girl from a group change over time?

Yes, the probability of choosing a girl from a group can change over time. This can be due to changes in the gender distribution of the group or changes in external factors that may affect the probability, such as preferences or biases.

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