# Probability of choosing a committee of 4 from 11 people given that 2 people must be a part?

As is obvious from the question, what is the probability of choosing a committee (through random selection) of 4 from a group 11 people given that person X and Y have to be a part of it?

What I did was ${\binom{2}{2}\binom{9}{2}\over \binom{11}{4}}$ however the answer the book gives is $\approx 0.655$. How did they get this answer?

Likewise another question I seem to get wrong is if 5 people sit randomly on a 5 seat row what is the probability that a) they sit at the ends of the row b) they sit together. I thought question a) was rather ambiguous, but as for b) I got $8\over 120$ which is also wrong. Does anyone know what I am doing wrong?

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Should I read the question as: a committee is chose at random from 11 people (using some process that you should specify). With the knowledge that $X$ and $Y$ have been in fact chosen on the committee, what is the probability that the committee has exactly $4$ members? I think not, but that is what your question suggests. – Marc van Leeuwen Jan 8 '12 at 13:10
@MarcvanLeeuwen you are correct. I will edit the question now. – E.O. Jan 8 '12 at 21:17

Your questions a a bit vague. For the first problem, I think you want "Find the probability that persons $X$ and $Y$ are on a randomly chosen 4 person committee selected from 11 people if $X$ and $Y$ are amongst the 11 people". Your answer is correct. (Note that it reduces to $6/55$; I imagine answer in the solution manual is in error.)

For the second question, do you want: "Tom and Bob are amongst 5 people. The five people randomly sit in a row..."

If so, then for part b:

The number of all seating arrangements is $5!=120$.

The number of arrangements in which Bob and Tom sit together is $8\cdot 6 = 48$. So, the required probability for part b) is $48/120$ (Imagine Bob and Tom taking seats first. There are 8 ways for them to do this. After Bob and Tom take their seats, the other three people can be seated in $3!$ different ways).

For part a), I'll assume that one of Tom and Bob sits at one end of the row, and the other at the other end of the row.

There are 2 ways for Tom and Bob to sit at the ends of the row; and, for each of these, there are $3!=6$ ways for the other three people to be seated. So, the required probability for part a) is $2\cdot 6\over120$.

You could solve part b) by thinking of Tom and Bob as "one" person with Tom to the left of Bob. There are $4!=24$ ways to seat four people; but for each particular seating of the four people, Tom and Bob can switch seats, giving a new seating. Thus the number of ways the five people can be seated with Tom next to Bob is $2\cdot 24=48$.

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Thanks for the answer! The first question had been bugging me for quite some time as I just couldn't seem to get it right despite the apparent simplicity of the question. As for the second question I appreciate the valuable insight. I suppose I had come to view the answer pages at the back of the book as always right and so when my answered differed I disregarded the possibility that the book might have been incorrect. – E.O. Jan 8 '12 at 12:30