# Conditional probability

#### Felicia

##### New member
A committee of 14 people consists of 4 boys, 5 girls and 5 other people are arranged in a row. Given that the boys are separated, find the probability of the girls are not all together

#### Jomo

##### Elite Member
Can any of the 5 other people be boys?

#### pka

##### Elite Member
A committee of 14 people consists of 4 boys, 5 girls and 5 other people are arranged in a row. Given that the boys are separated, find the probability of the girls are not all together
Of course other people must be either male or female. If the there are $$\displaystyle m$$ total males & $$\displaystyle f$$ total females in order for the males to be separated it has to be the case that $$\displaystyle 4\le m\le 7$$. Given that the males are separated there must be at least $$\displaystyle m-1$$ females.
Could there be a committee of these people in which $$\displaystyle m=8~\&~f=6~?$$ Well yes there could because $$\displaystyle m+f=14$$ but there would be no way to separate all of the males; there are only six separators that create seven separate spaces.
Now a word about the phrase "not all together" it means that at least two of the same sex are standing together.
Let's take an example: $$\displaystyle m=6~\&~f=8$$. If this example it takes five females to separate the six males but there are not enough males to separate the females. Anyone tackling this question needs to fully understand the setup.
There are $$\displaystyle 14!= 87,178,291,200$$ ways to arrange fourteen people in a row.
There are $$\displaystyle (8!)\left(^9\mathcal{C}_6\right)(6!)=2438553600$$ ways to arrange the committee so that no two of the six males are together.
Explanation: There are $$\displaystyle 8!$$ ways to arrange the females who create nine places to separate the six males who can be arranged in $$\displaystyle 6!$$ ways.
Thus if we are given six men and eight women to arrange in a row the probability that all the six men are separated is $$\displaystyle \frac{4}{143}$$ SEE HERE

Now the downer: I have no idea how to work this question. It is so poorly put as to be unintelligible.
1) How is the committee chosen? Are the first 4 boys & 5 girls fixed?
2) Even if they are or are not, how is the makeup of the other five people determined?
3) Was the complete question given?

I hope someone reading this or the author of the O.P. can help us here.

#### Dr.Peterson

##### Elite Member
I'll suppose that the "other people" are adults (or something else that isn't boys or girls), not that they are neither male nor female, or that they are just not known. The problem certainly could be clearer.

So you have some arrangement of BBBBGGGGGOOOOO. How many ways can they be arranged with no B's together? How many ways can they be arranged so that also GGGGG are all together? Where can you go from here?

There are several ways this could be tackled. In order to help as well as possible, we'll need to see what you have tried, and what you have learned. You did read our submission guidelines, didn't you?

#### pka

##### Elite Member
I'll suppose that the "other people" are adults (or something else that isn't boys or girls), not that they are neither male nor female, or that they are just not known. The problem certainly could be clearer.
So you have some arrangement of BBBBGGGGGOOOOO. How many ways can they be arranged with no B's together? How many ways can they be arranged so that also GGGGG are all together? Where can you go from here?
Well Prof. Peterson my initial reaction was "are you serious "?
Well O.K. say that the $$\displaystyle O's$$ are androgynous adults added to the given four boys and five girls (meaning prepubescent) children.
Now we have ten separators for the 'boys' so that can be done in $$\displaystyle (10!)\left(^{11}\mathcal{C}_4\right)(4!)=28740096000$$ ways.

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