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?
 A: 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 $2\cdot 24=48$.
