Expert: Josh - 8/19/2003

Question
A group of 6 women and 4 men
3 will be selected ramdomly
Find the probability that all will be men and that exactly 2 will be women.
I have the answer, but I need to know the steps in between.  

Answer
Hi Dolores,

Let's look at a picture to understand what's going on.
NOTATIONS:
I use P(X[i]=x) to denote the probability of event X[i]=x.

Case A: To pick 3 men.
Start with group {w,w,w,w,w,w,m,m,m,m}.
Pick 1: probability of selecting a man from 10 people is
P(X[1]=m)=4/10.
We then have {w,w,w,w,w,w,m,m,m}.
Pick 2: the conditional probability of selecting a man from this group is now 3 out of 9, given that the first pick was a male.
P(X[2]=m|X[1]=m)=3/9.
Next, we have {w,w,w,w,w,w,m,m}.
Pick 3: the conditional probability of selecting a man from this group is now 2 out of 8, given that the first and second pick were both male.
P(X[3]=m|X[1]=m,X[2]=m)=2/8.
Now, the probability of the composite event (all three happening) is the product of these probabilities. i.e.,
P(X[1]=m,X[2]=m,X[3]=m)
=P(X[1]=m)*P(X[2]=m|X[1]=m)*P(X[3]=m|X[1]=m,X[2]=m)
=(4/10)*(3/9)*(2/8)
=24/720

Notes: if a person of the same sex is replaced after each pick, in other words, the size of the group does not become diminishingly smaller (instead, remains constant), then the probability of each subsequent pick will be conditionally independent of earlier ones.
In that case, the conditional probability P(X[i]|X[i-1],X[i-2],...,X[1]) would be the same as the P(X[i]).

Can you apply the same logic for case B?
Hint: the order of selection does not matter, all we need is to end up with {m,w,w} or {w,m,w} or {w,w,m} in our selection.

In brief, P(X[i]=w,X[j]=w,X[k]=m)=(6/10)*(5/9)*(4/8).
You can choose any of the three permutations.

Cheers.