# Is this proof of indepencence of two events circular?

Task (I freely admit it is from a former test):

We choose a random person $X$ and it turns out that this person has three siblings, with at least one older sister among them.

A) What is the probability that $X$ is a girl?

B) What is the probability that the second oldest child of this familily is a girl?

C) What is the probability that $X$ is the youngest child of this family?

D) Let $A$, $B$, $C$ denote the above events. Which pairs are independent: $(A, B)$, $(A, C)$, $(B, C)$? Validate each answer.

Note: Assume that, regarding each birth, the events: a boy is born and a girl is born have the same probability $p=\frac12$ (we do not take into account the possibility of twins being born, etc), and these events are independent from any former births.

# My solution attempt:

A) $\frac12$, from the independence of birhts.

Before we go to B), C), D), let's list all legal combinations:

• If $X$ is the second oldest kid: $b/g\quad b/g\quad X_{b/g}\quad g$ - this gives $8$ combinations
• If $X$ is the third oldest kid: $b/g \quad X_{b/g}\quad b/g \quad b/g$ - this would be $4\cdot 4$ combinations, but we must exclude the illegal combination that two oldest kids are all boys, so it's $4\cdot 3 = 12$ combinations
• If $X$ is the youngest kid: $X_{b/g} \quad b/g \quad b/g \quad b/g$ - Excluding the illegal combination that all older kids are all boys this gives us $2\cdot7=14$ combinations

So we have a total of $34$ legal combinations.

B) Let's list legal combinations:

• $b/g\quad b/g\quad X_g \quad g$ - $4$ combinations
• $b/g \quad X_{b/g} \quad g \quad b/g$ - $8$ combinations
• $X_{b/g} \quad b/g \quad g \quad b/g$ - $8$ combinations

Thus the probability is $\frac{4+8+8}{34}=\frac{10}{17}$

C) Above we showed that there are $14$ legal combinations in such a case, so the answer is $\frac{14}{34}=\frac{7}{17}$

D) And here be dragons, I guess.

This is how I'd tackle this:

$(A, B)$:

All legal combinations for the event $A\cap B$ are:

• $b/g\quad b/g \quad X_g \quad g$ - $4$ combinations
• $b/g \quad X_g \quad g \quad b/g$ - $4$ combinations
• $X_g\quad b/g\quad g\quad b/g$ - $4$ combinations

$12$ combinations total, so $P(A\cap B) = \frac {12}{34} \neq \frac12\cdot \frac{20}{34} = P(A)\cdot P(B)$, so by definition $A$ and $B$ are not independent.

$(A,C)$: There are $7$ legal combinations for the event $A\cap C$: $X_g\quad b/g \quad b/g \quad b/g$, this would be $8$ but we must exclude the illegal combination $X_g \quad b \quad b \quad b$

Therefore $P(A\cap C) = \frac7{34} = \frac12 \cdot \frac {14}{34} = P(A)\cdot P(B)$, so by definition these events are independent

$(B, C)$: THere are $8$ legal combinations: $X_{b/g} \quad b/g \quad g \quad b/g$

So we have $P(B\cap C) = \frac8{34} = \frac 4{17}$ But $P(B)\cdot P(C) = \frac7{17}\cdot \frac{10}{17} \neq \frac 4{17}$

So by definition these events are not independent.

# My worry is:

I used the classical definition of probability for each part of the task. But I can only use the classical definition if the permutation of children in this famility is independent of their sex; so, essentialy, I can only use the classical definition if $(A, C)$ are independent. However, it was my task to prove that $A$ and $C$ are really independent, and I used the classical definition to prove it... So, in the end, am I not guilty of circular reasoning here?

The reason this is okay and not circular is the subtlety of how person $X$ is selected, and what information we have available about them.
When you say that $X$ is selected a random, presumably you mean that they are chosen among the four children uniformly and independently of the sex of the children. If we didn't know anything about the older siblings of $X$, then you're right; $A$ and $C$ would be trivially independent of one another because of that assumption. Note that in this scenario, $(A, B)$ and $(B, C)$ would also be independent for essentially this same trivial reason.
But in this case, we have some extra information -- call it event $E$ -- that $X$ has an older sister. Now, the problem is secretly to compute $P(A \mid E), P(B \mid E), P(C \mid E)$ and to assess their independence on one another. The computation you actually did was $P(A \cap C \mid E) = P(A \mid E) \cdot P(C \mid E)$. If this was indeed just a trivial application of the principle that the permutation is independent of the sex, then I claim that the other events would have been independent as well. Instead, what you did was an explicit calculation on the conditioned probability space (this was the act where you removed illegal combinations from consideration) and verified the conditional independence.
I think where you're getting hung up is that you have a strong intuitive sense that $A$ and $C$ should remain independent of one another even with the additional information provided. And you're right, of course; but you didn't rely on that intuition to show it. Instead, you made a computation to verify it.