Chapter 13 · Probability

Miscellaneous · Q12 (MCQ): $P(A\mid B) > P(A)$ gives $P(B\mid A) > P(B)$ ⁠, option (C)

Miscellaneous ExerciseConditional probability · Section 13.2

The question

Choose the correct answer.

If $P(A|B) > P(A)$ ⁠, then which of the following is correct :

(A) $P(B|A) < P(B)$ (B) $P(A \cap B) < P(A) \cdot P(B)$

Exam answer

\begin{aligned} &P(A|B) > P(A) \\ \Rightarrow\ &\frac{P(A \cap B)}{P(B)} > P(A) \\ \Rightarrow\ &P(A \cap B) > P(A)\,P(B) \end{aligned}

So $P(A \cap B) > 0$ ⁠, hence $P(A) \neq 0$ ⁠. Dividing by $P(A)$ ⁠,

\begin{aligned} &\frac{P(A \cap B)}{P(A)} > P(B) \\ \Rightarrow\ &P(B|A) > P(B) \end{aligned}

Answer: (C) $P(B|A) > P(B)$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Both $P(A|B) > P(A)$ and $P(B|A) > P(B)$ say the same thing in a different form: $P(A \cap B) > P(A)\,P(B)$ ⁠, that is, $A$ and $B$ happen together more often than independent events would. That statement treats $A$ and $B$ alike, so if knowing $B$ makes $A$ more likely, knowing $A$ makes $B$ more likely.

Given, to find, and the plan

Given: Events $A$ and $B$ with $P(A|B) > P(A)$ ⁠. For $P(A|B)$ to exist, $P(B) \neq 0$ ⁠.
To find: Which of the four statements must be true.
Plan: (1) Turn the given inequality into one about $P(A \cap B)$ ⁠. (2) Show $P(A) \neq 0$ ⁠. (3) Divide by $P(A)$ ⁠. (4) Test every option on a die.

Basics you need

1Conditional probability (Section 13.2)

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ for $P(B) \neq 0$ ⁠, and $P(B|A) = \frac{P(A \cap B)}{P(A)}$ for $P(A) \neq 0$ ⁠. Lesson: conditional probability

2Independent events (Section 13.4)

$A$ and $B$ are independent when $P(A \cap B) = P(A)\,P(B)$ ⁠, equivalently $P(A|B) = P(A)$ ⁠. Lesson: independent events

3Inequalities and positive numbers

Multiplying or dividing both sides of an inequality by a positive number keeps its direction. A probability is never negative.

Solution

Rewrite the given inequality

$P(A|B)$ is defined, so $P(B) > 0$ ⁠. Then
$\begin{aligned} &P(A|B) > P(A) \\ \Rightarrow\ &\frac{P(A \cap B)}{P(B)} > P(A) \\ \Rightarrow\ &P(A \cap B) > P(A)\,P(B). \end{aligned}$
Multiplying both sides by the positive number $P(B)$ keeps the direction of " $>$ ". The result no longer has a fraction, and it treats $A$ and $B$ in the same way.

Section 13.2, Definition 1. Lesson: conditional probability
$P(A)$ is not 0

$P(A)\,P(B) \ge 0$ ⁠, so Step 1 gives $P(A \cap B) > 0$ ⁠. Since $A \cap B$ is part of $A$ ⁠, $P(A) \ge P(A \cap B) > 0$ ⁠.

We want to divide by $P(A)$ next, and $P(B|A)$ is only defined when $P(A) \neq 0$ ⁠. The question does not say so, but it follows from the given inequality.

$P(E) \le P(F)$ when $E \subset F$ ⁠. Basics: rules of probability
Divide by $P(A)$

Dividing both sides of $P(A \cap B) > P(A)\,P(B)$ by the positive number $P(A)$ ⁠:
$\begin{aligned} &\frac{P(A \cap B)}{P(A)} > P(B), \\ &\text{that is, } P(B|A) > P(B). \end{aligned}$
This is option (C). The steps also run backwards (when $P(A), P(B) \neq 0$ ⁠), so $P(A|B) > P(A)$ and $P(B|A) > P(B)$ are equivalent.

$P(B|A) = \frac{P(A \cap B)}{P(A)}$ ⁠. Lesson: conditional probability
Examine every option

Throw a fair die and take $A = \{1, 2, 3\}$ (a number at most 3) and $B = \{1, 2\}$ ⁠. Then $P(A) = \frac12$ ⁠, $P(B) = \frac13$ ⁠, $P(A \cap B) = \frac13$ ⁠, and
$\begin{aligned} P(A|B) &= \frac13 \div \frac13 = 1, \\ P(B|A) &= \frac13 \div \frac12 = \frac23. \end{aligned}$
So $P(A|B) = 1 > \frac12 = P(A)$ ⁠: this pair satisfies the question.
- (A) $P(B|A) < P(B)$ ⁠. Here $\frac23 < \frac13$ is false; Step 3 shows the opposite always holds. Wrong.
- (B) $P(A \cap B) < P(A) \cdot P(B)$ ⁠. Here $\frac13 < \frac16$ is false; Step 1 shows the opposite always holds. Wrong.
- (C) $P(B|A) > P(B)$ ⁠. $\frac23 > \frac13$ ⁠, and Step 3 proves it in general. Correct.
- (D) $P(B|A) = P(B)$ ⁠. That would mean $A$ and $B$ are independent, and then $P(A|B) = P(A)$ ⁠, not $>$ ⁠. Wrong.
The die example shows that (A), (B) and (D) can fail; Step 3 shows that (C) cannot.

Independence: $P(A|B) = P(A)$ (Section 13.4, Definition 2). Lesson: independent events

Result

$P(A|B) > P(A)$ gives $P(A \cap B) > P(A)\,P(B)$ ⁠, and so $P(B|A) > P(B)$ ⁠: option (C). (Textbook answer: C.)

Check it with numbers

In the die example, $P(A)\,P(B) = \frac12 \times \frac13 = \frac16 \approx 0.1667$ ⁠, less than $P(A \cap B) = \frac13 \approx 0.3333$ ⁠. The verify script takes every pair of events on a die with $P(A)$ and $P(B)$ not zero and $P(A|B) > P(A)$ ⁠: 1562 pairs. (C) holds in all of them; (A), (B) and (D) in none.

Common mistakes

Choosing (A). Conditioning does not "use up" the link: the effect goes the same way in both directions.
Choosing (D). $P(B|A) = P(B)$ is the independent case, which the strict " $>$ " rules out.
Dividing by $P(A)$ without checking it is not 0. Step 2 shows it is positive here.

What to take away

$P(A|B) > P(A)$ ⁠, $P(B|A) > P(B)$ and $P(A \cap B) > P(A)\,P(B)$ say the same thing.
Likewise " $<$ " in one form gives " $<$ " in the others, and " $=$ " is independence.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \neq 0$	Steps 1, 3, 4	Section 13.2, Definition 1 (Lesson: conditional probability)
$0 \le P(E) \le P(F)$ for $E \subset F$	Step 2	Class XI axioms (Basics: rules of probability)
Independence: $P(A\|B) = P(A)$	Step 4	Section 13.4, Definition 2 (Lesson: independent events)
$P(E) = \frac{n(E)}{n(S)}$ on a fair die	Step 4	Classical probability, Class XI (Basics: classical probability)

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Miscellaneous · Q12 (MCQ): P(A∣B)>P(A)P(A\mid B) > P(A)P(A∣B)>P(A) gives P(B∣A)>P(B)P(B\mid A) > P(B)P(B∣A)>P(B)⁠, option (C)

Basics you need

Solution

Rewrite the given inequality

P(A)P(A)P(A) is not 0

Divide by P(A)P(A)P(A)

Examine every option

Every fact used, and where it comes from

Miscellaneous · Q12 (MCQ): $P(A\mid B) > P(A)$ gives $P(B\mid A) > P(B)$ ⁠, option (C)

$P(A)$ is not 0

Divide by $P(A)$