I'm just floating this and I'm not sure how to describe my problem precisely. So I'll lead with an example:
Assume a team of 3 are chosen from 4 boys and 5 girls. Let the rv $X$ be the number of girls on the team. It follows that $X=0,1,2,3$. In constructing the pdf of $X$, my initial impulse was to take into account the inter-dependence of choosing 3 students: i.e. the outcomes-girl or boy-of each choice except the first depends on the preceding outcome(s). This is the probability tree I drew to illustrate the sample space of this problem:
I've highlighted the patterns I'm talking about in red and green asterisks. For example, although the sequences (if you visualize them as such) of choosing a girl, another girl and a boy ($GGB$), or a girl, a boy and another girl ($GBG$), are different, their joint probabilities are equal (i.e. $P(GGB)=P(GBG)=P(BGG)$). As far as my problem (constructing the pdf of $X$) is concerned, I can work out the probability of just one of these sequences, then conjecture that there are $\frac{3!}{2!}$ ways of choosing two girls and a boy in a sequence and multiply the probability by $\frac{3!}{2!}$ to get $P(X=2)$ (like I've done below the probability tree for $P(X=1)$ and $P(X=2)$).
Is there a mathematical basis for this pattern or am I reading too much into it? (FYI I'm new to probability theory).
Best Answer
What you're seeing will happen in general. Look at the numbers you've put red stars on. They have numerators $4 \times 3 \times 5$ (BBG), $4 \times 5 \times 3$ (BGB), and $5 \times 4 \times 3$ (GBB). Those are the same number. Consider just the numerator - how many ways can you pick a boy, another boy, and finally a girl? There are four ways to pick the first boy, three ways to pick the second boy, and then five ways to pick the girl. But if you rearrange and pick, say, a boy, a girl, and then another boy, you just get the same factors in a different order.