Trouble describing correctly a conditional probability problem.

Question

I have the following problem : We have 55% of students who attend a class that comes from school X and the rest is from school Y. From school X 80% are bilingual. From school Y only 50% are bilingual.

What is the probability of having a bilingual student that comes from school Y?

This seems to be described by $P(SchoolY|Bilingual)$. We know that $P(SchoolY) = 0.45$ and we know that $P(Bilingual$ $\cap$ $ SchoolY )$ = 0.50.

Is it correct to determine that $P(SchoolY|Bilingual) = $ $P(Bilingual$ $\cap$ $ SchoolY )$ $\times$ $P(SchoolY) = 0.225$ ?

Answer 1

To answer your question:

Is it correct to determine that $P(SchoolY|Bilingual) = P(Bilingual \cap SchoolY)\times P(SchoolY)$?

No. By definition of the conditional probability, that expression should be: $$P(SchoolY|Bilingual) = \frac{P(Bilingual \cap SchoolY)}{P(Bilingual)}$$

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Update: From a comment by OP, the exact problem statement is:

A student is randomly picked from the group. Consider now that the student is bilingual.What is the probability that the student is from school Y?

We are given these: $$P(school=X)=0.55\qquad P(school=Y)=0.45$$ $$P(bilingual\;|\;school=X)=0.8\qquad P(bilingual\;|\;school=Y)=0.5$$

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We need the quantity $P(school=Y\;|\;bilingual)$. By Bayes' theorem it will be : $$\begin{equation}P(school=Y\;|\;bilingual) = P(bilingual\;|\;school=Y)\cdot \frac{P(school=Y)}{P(bilingual)}\\ \;\;\;=\frac{0.225}{P(bilingual)}\end{equation}$$ Similarly, we can calculate, $P(school=X\;|\;bilingual)$ as: $$\begin{equation}P(school=X\;|\;bilingual) = P(bilingual\;|\;school=X)\cdot \frac{P(school=X)}{P(bilingual)}\\ \;\;\;=\frac{0.44}{P(bilingual)}\end{equation}$$ But given a student is bilingual she can only be from one of these schools, so the sum of these is 1. Thus: $$\frac{0.225}{P(bilingual)}+\frac{0.44}{P(bilingual)}=1\implies P(bilingual)= 0.225+0.44= 0.665$$ And then, we have: $$P(school=Y\;|\;bilingual)=\frac{0.225}{0.665}\approx0.338$$ $$P(school=X\;|\;bilingual)=\frac{0.44}{0.665}\approx0.662$$

Answer 2

Assuming the percentages are meaningful--that is, there actually are students in this class--the probability of having a bilingual student that comes from school Y is $1.$ We know that $25$% of students come from school Y and are bilingual; but if there were not at least one bilingual student from school Y, that percentage would have been $0$% instead of $25$%.

The way we get probabilities that are not $0$ or $1$ is typically that we look at the outcome of some random procedure. For example you choose a student at random from the class.

If we then want to talk about the probability that a randomly selected student is a bilingual student that comes from school Y, this is $P(\text{bilingual} \cap \text{SchoolY})$. That is, the randomly selected student is bilingual and is from school Y.

If we then want to talk about the probability that a randomly selected student is a bilingual student, given that this student comes from school Y, this is $P(\text{bilingual} \mid \text{SchoolY})$. That is, we are asking if the student comes from school Y, then what would the probability be that the student is bilingual.

The formula $P(\text{SchoolY} \mid \text{bilingual})$ is the probability that a randomly selected student is from school Y, given that the student is bilingual.

It is hard for me to imagine how "the probability of having a bilingual student that comes from school Y" can be interpreted as $P(\text{SchoolY} \mid \text{bilingual})$, even when the random selection of the student is specified.

But let's consider the following question:

A student is randomly picked from the group. Consider now that the student is bilingual. What is the probability that the student is from school Y?

This is asking for the probability that a randomly selected student is from school Y, given that the student is bilingual: $P(\text{SchoolY} \mid \text{bilingual})$. It is the order of the sentences that matters: first that a student is randomly selected; second, we find out (or are told to assume) that the student is bilingual; and third, after the information in the previous sentences is given, we want the probability the student is from school Y.

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Regarding your conclusions, from the given information we know that $$P(\text{bilingual} \cap \text{SchoolY}) \neq 0.50.$$

Also, except in some special cases (which do not apply here), $$P(Y\mid \text{bilingual}) \neq P(\text{bilingual} \cap \text{SchoolY}) \times P(\text{SchoolY}).$$ I recommend you review the formulas of conditional probability to see why these two formulas are not equal, and what correct equation you could have written instead.