Exercise 13.3 · Q14 (MCQ): $A \subset B$⁠: $P(A\mid B) \ge P(A)$⁠, option (C)

Basics you need

Solution

1. What $A \subset B$ tells us

   Every outcome of $A$ is also an outcome of $B$⁠, so the outcomes in both $A$ and $B$ are just the outcomes of $A$⁠:

   $$A \cap B = A.$$

   Also $P(A) \le P(B)$⁠, because $B$ contains all of $A$ and possibly more.

   The definition of $P(A|B)$ needs $P(A \cap B)$⁠. For a subset, the intersection is the smaller set itself, so $P(A \cap B) = P(A)$⁠.
   Events are sets of outcomes; $A \subset B$ means $A \cap B = A$⁠. Basics: events as sets

2. Write $P(A|B)$ by the definition

   Since $P(B) \neq 0$⁠, the definition of conditional probability applies:

   $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}.$$
   The condition $P(B) \neq 0$ in the question is exactly what the definition needs: we may not divide by 0.
   $P(E|F) = \frac{P(E \cap F)}{P(F)}$⁠, $P(F) \neq 0$ (Section 13.2, Definition 1). Lesson: conditional probability

3. Compare with $P(A)$

   We know $0 < P(B) \le 1$ and $P(A) \ge 0$⁠. Dividing a number that is not negative by a positive number at most 1 cannot make it smaller, so

   $$\begin{aligned} &\frac{P(A)}{P(B)} \ge P(A), \\ &\text{that is, } P(A|B) \ge P(A). \end{aligned}$$

   In detail: $\frac{P(A)}{P(B)} - P(A) = P(A) \cdot \frac{1 - P(B)}{P(B)}$⁠, a product of numbers that are not negative.

   The difference is 0 exactly when $P(A) = 0$ or $P(B) = 1$⁠, so the inequality can be an equality. That is why the correct statement has $\ge$ and not $>$⁠.
   $0 \le P(E) \le 1$ for every event. Basics: rules of probability

4. Examine every option

   Take one throw of a fair die, $S = \{1, 2, 3, 4, 5, 6\}$⁠, with $B = \{1, 2, 3, 4\}$ and $A = \{1, 2\}$⁠, so $A \subset B$⁠. Then $P(A) = \frac26 = \frac13$⁠, $P(B) = \frac46 = \frac23$ and

   $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac13 \div \frac23 = \frac12.$$
   • (A) $\frac{P(B)}{P(A)} = 2$⁠, but $P(A|B) = \frac12$⁠. In general $\frac{P(B)}{P(A)} \ge 1$ (when $P(A) > 0$⁠) while a probability is at most 1, so (A) can hold only when $P(A) = P(B)$⁠, and it means nothing when $P(A) = 0$⁠. Wrong.
   • (B) $\frac12 < \frac13$ is false, and Step 3 shows $P(A|B) < P(A)$ never happens. Wrong.
   • (C) $\frac12 \ge \frac13$ is true, and Step 3 proves it for every pair $A \subset B$⁠. Correct.
   • (D) Since (C) is always correct, "None of these" is wrong.

   The equality case: with $B = S$⁠, $P(B) = 1$ and $P(A|S) = P(A)$⁠, so "$>$" would be false there. The sign $\ge$ covers it.

   One example cannot prove (C), but Step 3 does that; the example only shows that (A) and (B) fail. An option that is not always true is not "correct".
   Knowing $B$ shrinks the sample space to $B$⁠. Examples 1 to 7

Every fact used, and where it comes from