Exercise 13.1 · Q17 (MCQ): Events with $P(A|B) = P(B|A)$

Basics you need

Solution

1. Write both sides with the definition

   $$P(A|B) = \frac{P(A\cap B)}{P(B)},$$ $$P(B|A) = \frac{P(A\cap B)}{P(A)}.$$
   For $P(B|A)$ the definition gives $\frac{P(B\cap A)}{P(A)}$⁠, and $B\cap A = A\cap B$⁠. This is how the book itself rewrites $P(F|E)$ at the start of Section 13.3. So the condition $P(A|B) = P(B|A)$ reads $\frac{P(A\cap B)}{P(B)} = \frac{P(A\cap B)}{P(A)}$⁠.
   Definition 1, Section 13.2. Lesson: conditional probability

2. Cancel the common numerator

   If $P(A\cap B) \ne 0$⁠, divide both sides by it:

   $$\frac{1}{P(B)} = \frac{1}{P(A)}\quad\Rightarrow\quad P(A) = P(B).$$
   Dividing by a non-zero number keeps an equation true; then taking reciprocals of both sides (both are non-zero) gives $P(A) = P(B)$⁠. So statement (D) follows from the condition.
   Equal fractions with the same non-zero top have equal bottoms. Basics: working with fractions

3. Test the options on a die

   Roll a fair die and take $A = \{1, 2\}$⁠, $B = \{2, 3\}$⁠. Then $A\cap B = \{2\}$⁠, and

   $$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} = \frac{1/6}{2/6} = \frac{1}{2}, \\ P(B|A) &= \frac{P(A\cap B)}{P(A)} = \frac{1/6}{2/6} = \frac{1}{2}. \end{aligned}$$
   The condition holds, so every statement that "must" follow has to be true here. As step 2 predicts, $P(A) = \frac26 = P(B)$⁠.
   Classical probability on a die. Basics: P(E) = n(E)/n(S)

4. Examine every option

   (A) $A \subset B$ but $A \ne B$⁠. In the example, $1 \in A$ but $1 \notin B$⁠, so $A$ is not a subset of $B$⁠. Not forced. (In fact for any events $X \subset Y$ with $P(X) \ne 0$ we get $P(Y|X) = 1$⁠, while $P(X|Y) = \frac{P(X)}{P(Y)}$⁠.)

   (B) $A = B$⁠. If $A = B$⁠, both conditional probabilities are 1, so the condition holds. But the example has $A \ne B$ and still satisfies it. Not forced.

   (C) $A\cap B = \phi$⁠. The example has $A\cap B = \{2\}$⁠, not empty. Not forced.

   (D) $P(A) = P(B)$⁠. Proved in step 2, and true in the example. Correct.

   (B) and (C) are each one way to make the condition true, but the question asks what the condition guarantees. Only (D) holds every time the common value is non-zero.
   A counterexample rules an option out. Basics: events as sets

5. The special case $P(A\cap B) = 0$

   Step 2 divided by $P(A\cap B)$⁠. If it is 0, both conditional probabilities are 0 and the condition holds automatically. For example, $C = \{1\}$ and $D = \{2, 3\}$ on the die give $P(C|D) = 0 = P(D|C)$⁠, although $P(C) = \frac16$ and $P(D) = \frac26$⁠.

   So, strictly, (D) follows only when the common value of $P(A|B)$ and $P(B|A)$ is not 0, that is when $P(A\cap B) \ne 0$⁠. The question, and the answer key's (D), assume this without saying so. Without that assumption no option is guaranteed: the pair $C$⁠, $D$ above breaks (D), and the die example of Step 3 breaks (A), (B) and (C). With it, (D) is guaranteed and the other three are not, so (D) is the intended answer.
   Disjoint events have $P(E\cap F) = 0$⁠. Basics: two events cut S into four regions

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