# Conditional Probability Question

• GreenPrint
In summary, the conversation is about calculating the probability of selecting a group of 4 people with an equal number of men and women from a population of 2n individuals, using the formula P=(A|B) = \frac{P(A \bigcap B)}{P(B)}. The simplified answer is [C(n,2)*c(n,2)]/C(2n,4).
GreenPrint

## Homework Statement

A population has n men and n women. If you where to take 4 people out of the population to form a group what's the probability that there are exactly the same number of men as women in the group .

## The Attempt at a Solution

Ok so I thought of this to some extent and am lost.

The population size is 2n

(2n)(2n-1)(2n-2)(2n-3) different possible groups if you consider each person of the population to be distinct people without regards to the condition of gender.

When you take gender into consideration there are

2*2*2*2 = 2^4 = 16 different combinations when take the gender condition into account and consider each individual to be distinct (that is GGBB is different from BBGG)

If I consider each man and women to be the same there are a total of 5 different combinations...
BBBB
BBBG
BBGG
BGGG
GGGG

I know the following formula

P=(A|B) = $\frac{P(A \bigcap B)}{P(B)}$

I'm assuming that I need to use this formula. The only problem is that I don't know how to come up with A and B.

Thanks for any help that nay one can provide.

GreenPrint said:

## Homework Statement

A population has n men and n women. If you where to take 4 people out of the population to form a group what's the probability that there are exactly the same number of men as women in the group .

## The Attempt at a Solution

Ok so I thought of this to some extent and am lost.

The population size is 2n

(2n)(2n-1)(2n-2)(2n-3) different possible groups if you consider each person of the population to be distinct people without regards to the condition of gender.

When you take gender into consideration there are

2*2*2*2 = 2^4 = 16 different combinations when take the gender condition into account and consider each individual to be distinct (that is GGBB is different from BBGG)

If I consider each man and women to be the same there are a total of 5 different combinations...
BBBB
BBBG
BBGG
BGGG
GGGG

I know the following formula

P=(A|B) = $\frac{P(A \bigcap B)}{P(B)}$

I'm assuming that I need to use this formula. The only problem is that I don't know how to come up with A and B.

Thanks for any help that nay one can provide.

It's just the number of ways to pick 2 women from n women times the number of ways to pick 2 men from n men divided by the number of ways to pick 4 people from 2n people regardless of gender, isn't it? You seem to be really over complicating this.

Dick said:
It's just the number of ways to pick 2 women from n women times the number of ways to pick 2 men from n men divided by the number of ways to pick 4 people from 2n people regardless of gender, isn't it? You seem to be really over complicating this.

$(N_{K})^{2}$

Square N choose K in this notation? I have never seen this before don't see why I can't use this notation.

I couldn't figure out how to do this in itex code

GreenPrint said:

$(N_{K})^{2}$

Square N choose K in this notation? I have never seen this before don't see why I can't use this notation.

I couldn't figure out how to do this in itex code

I'm not exactly sure what the jist of your question is, but without even using tex, if C(n,k) is the number of ways to choose k objects from n, then what's your answer? And then write it explicitly using a formula for C(n,k).

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

GreenPrint said:

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

I would write it as C(n,2)^2. But that's not terribly important. And the denominator should be C(2n,4), yes? You know C(n,k)=n!/(k!(n-k)!), right? I would guess they want you to use that to simplify [C(n,2)*C(n,2)]/C(2n,4).

GreenPrint said:

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

If your denominator were C(2n,4) then, yes, your answer would be correct. This is just a so-called hypergeometric probability.

## 1. What is conditional probability?

Conditional probability is a measure of the likelihood of an event occurring, given that another event has already occurred. It takes into account the relationship between two events and is expressed as P(A|B), where A and B are events.

## 2. How is conditional probability calculated?

Conditional probability is calculated by dividing the probability of the intersection of the two events by the probability of the first event. This can be written as P(A|B) = P(A ∩ B) / P(B).

## 3. What is the difference between conditional probability and joint probability?

Conditional probability only takes into account the probability of an event occurring given that another event has occurred, while joint probability considers the probability of both events occurring together. Conditional probability is a subset of joint probability.

## 4. What is the relationship between conditional probability and independence?

If two events are independent, then the conditional probability of one event occurring given the other event has occurred is equal to the unconditional probability of the first event. This means that the occurrence of one event does not affect the probability of the other event.

## 5. How is conditional probability used in real-life situations?

Conditional probability is used in many real-life situations, such as in weather forecasting, medical diagnoses, and risk assessment. It allows us to make more accurate predictions and decisions by taking into account the relationship between events.

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