Finding probability function (simple problem)

Question

I have the following problem with solution, but I do not understand the solution. I will appreciate your help.

Problem: In class there are 10 pupils: 3 girls and 7 boys. Every morning the teacher randomly allocates the pupils to pairs. What is probability function X for the number of mixed (boy + girl) couples?

Solution:

[Stage 1] Number of mixed pairs can be either 3 or 1.

[Stage 2] For calculating probability we need to check the gender of the partner of each one of the girls.

[Stage 3] {X=3} will happen if for each girl will be a boy as partner. $P(X=3)=\dfrac79 \times\dfrac67 \times 1 = \dfrac23$

[Stage 4] {X=1} might happen in two ways: a) the partner of girl No 1 is boy (7 possibilities) and partner of girl No 2 is also girl (1 possibility); b) the partner of girl No 1 is girl (2 possibilities) and for remaining girl we have a boy (no choice). $P(X=1) = \dfrac79 \times \dfrac17 + \dfrac29 \times 1 = \dfrac13$

What do I miss: In [stage 3] we check for partner for the first girl, then for the second and third. But where are the permutations of girls? For the pairing with $\dfrac79$ it can be any of three girls. Where is it taken into account? The same for [stage 4]: to start they talk about girl No 1, but it could be also girl No 2 or 3. Where is it taken into account?

Thanks a lot!

Answer 1

You could include permutations, but then you also have to multiply with the probability on that permutation. For stage 3, you get $$\frac{1}{3}\left(\frac{7}{9}\frac{6}{7}1\right) + \frac{1}{3}\left(\frac{7}{9}\frac{6}{7}1\right) + \frac{1}{3}\left(\frac{7}{9}\frac{6}{7}1\right) = \frac{7}{9}\frac{6}{7}.$$ This is different from when you are counting cases.

Answer 2

As another approach, you could note that the chances of Alice and Betty being paired is $1/9$, since each of the 9 other students has an equal chance of being Alice's partner. Similarly for Alice and Cathy, and also for Betty and Cathy. Since these three events are pairwise exclusive, the probability of one of them happening is just their sum $1/9 + 1/9 + 1/9 = 1/3$, and the complement (all three women paired with men) has probability $1-1/3=2/3$.

Answer 3

Let $B_1$ be the event that girl number 1 is paired with a boy, $B_2$ be the event that girl number 2 is paired with a boy, $B_3$ be the event that girl number 3 is paired with a boy. Then $$ P(X = 3) = P(B_1 \cap B_2 \cap B_3). \tag1 $$ Is there any doubt about Equation $(1)$? On the left we have the probability of an event $X = 3,$ which says there are three boy-girl pairings. On the right we have an event $B_1 \cap B_2 \cap B_3,$ which says that each girl is paired with a boy.

Is it possible for $B_1,$ $B_2,$ and $B_3$ all to be true and yet $X \neq 3$? Is it possible that $X = 3$ and yet there is any one or more of the events $B_1,$ $B_2,$ and $B_3$ that does not occur? No, we have three girl-boy pairings if and only if each girl is paired with a boy. The two events are identical, and they have the same probability.

Next we may observe that $$ P(B_1 \cap B_2 \cap B_3) = P(B_1) \, P(B_2 \mid B_1) \, P(B_3 \mid B_1 \cap B_2). \tag2 $$ Equation $(2)$ is a universal equation about conditional probability. It is true for any three events in any probability space, provided that all four of the probabilities in the equation are defined.

Now observe that \begin{align} P(B_1) &= \frac79, \\ P(B_2 \mid B_1) &= \frac67, \\ P(B_3 \mid B_1 \cap B_2) &= 1. \end{align}

Note that even though each girl is distinctly identified, so that permutations of them are distinguishable, the formulas are all still true. In fact these formulas do not even make sense unless you identify each girl distinctly from the others.

Now, if you like, you can permute the girls within the formulas. For example, we have $$ P(B_1 \cap B_2 \cap B_3) = P(B_2) \, P(B_3 \mid B_2) \, P(B_1 \mid B_2 \cap B_3). $$ But this is not a separate case from Equation $(2)$; both equations are describing the exact same event, $B_1 \cap B_2 \cap B_3.$ Even if you permute the girls within that event's description, for example if you write $B_2 \cap B_3 \cap B_1$ it is still the same event, not a new event to be counted.