Chapter 13 · Probability

Miscellaneous · Q13 (MCQ): $P(A) + P(B) - P(A \cap B) = P(A)$ gives $P(A\mid B) = 1$ ⁠, option (B)

Miscellaneous ExerciseConditional probability · Section 13.2

The question

Choose the correct answer.

If A and B are any two events such that $P(A) + P(B) - P(A \text{ and } B) = P(A)$ ⁠, then

(A) $P(B|A) = 1$ (B) $P(A|B) = 1$

Exam answer

$P(A) + P(B) - P(A \cap B) = P(A)$ gives $P(A \cap B) = P(B)$ ⁠, with $P(B) \neq 0$ for $P(A|B)$ to be defined.

\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{P(B)}{P(B)} = 1 \end{aligned}

Answer: (B) $P(A|B) = 1$

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

The one idea this problem is built on

The left side, $P(A) + P(B) - P(A \cap B)$ ⁠, is $P(A \cup B)$ ⁠. So the condition says that adding $B$ to $A$ adds no probability: $B$ lies inside $A$ ⁠. Then knowing that $B$ has happened makes $A$ certain, $P(A|B) = 1$ ⁠.

Given, to find, and the plan

Given: Events $A$ and $B$ with $P(A) + P(B) - P(A \text{ and } B) = P(A)$ ⁠, where "A and B" is the event $A \cap B$ ⁠.
To find: Which of the four statements must be true.
Plan: (1) Simplify the condition to $P(A \cap B) = P(B)$ ⁠. (2) Use the definition of $P(A|B)$ ⁠. (3) Read the result as a statement about sets. (4) Test every option on a die.

Basics you need

1The addition rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (Class XI). Basics: the addition rule

2Conditional probability (Section 13.2)

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

3Splitting an event

$B = (A \cap B) \cup (A' \cap B)$ ⁠, two disjoint pieces. Basics: Venn diagrams

Solution

Simplify the condition

"A and B" is the event $A \cap B$ ⁠. Subtract $P(A)$ from both sides of the given equation:
$\begin{aligned} P(A) + P(B) - P(A \cap B) &= P(A) \\ P(B) - P(A \cap B) &= 0 \\ P(A \cap B) &= P(B). \end{aligned}$
The left side of the given equation is $P(A \cup B)$ by the addition rule, so the condition is $P(A \cup B) = P(A)$ ⁠. Either way it comes down to $P(A \cap B) = P(B)$ ⁠.

Addition rule (Class XI). Basics: the addition rule
Apply the definition of $P(A|B)$

Options (B) and (D) use $P(A|B)$ ⁠, which is defined only when $P(B) \neq 0$ ⁠. Then
$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{P(B)}{P(B)} = 1. \end{aligned}$
The overlap of $A$ and $B$ has the same probability as $B$ itself, so the share of $B$ that lies in $A$ is all of it.

Section 13.2, Definition 1. Lesson: conditional probability
What it means for the sets

Since $B = (A \cap B) \cup (A' \cap B)$ with disjoint pieces,
$P(A' \cap B) = P(B) - P(A \cap B) = 0.$
With equally likely outcomes this means $A' \cap B = \phi$ ⁠: every outcome of $B$ is in $A$ ⁠, so $B \subset A$ ⁠.

The condition puts $B$ inside $A$ ⁠: then $A \cup B = A$ and $A \cap B = B$ ⁠, and given $B$ ⁠, the event $A$ is certain.

This is the situation of Question 11 with the roles of $A$ and $B$ swapped: $P(A|B) = 1$ goes with $B \subset A$ ⁠.

$A' \cap B = \phi$ exactly when $B \subset A$ ⁠. Basics: Venn diagrams
Examine every option

Throw a fair die and take $A = \{1, 2, 3\}$ and $B = \{1\}$ ⁠. Then $P(A) = \frac12$ ⁠, $P(B) = \frac16$ and $P(A \cap B) = \frac16$ ⁠, so
$\begin{aligned} &P(A) + P(B) - P(A \cap B) \\ &= \frac12 + \frac16 - \frac16 = \frac12, \end{aligned}$
which equals $P(A)$ ⁠: the pair satisfies the condition. Also $P(B|A) = \frac16 \div \frac12 = \frac13$ ⁠.
- (A) $P(B|A) = 1$ ⁠. Here $P(B|A) = \frac13$ ⁠. Not always true.
- (B) $P(A|B) = 1$ ⁠. $P(A|B) = \frac16 \div \frac16 = 1$ ⁠, and Step 2 proves it in general. Correct.
- (C) $P(B|A) = 0$ ⁠. Here $P(B|A) = \frac13$ ⁠. In general $P(B|A) = \frac{P(B)}{P(A)}$ ⁠, which is not 0 when $P(B) \neq 0$ ⁠. Wrong.
- (D) $P(A|B) = 0$ ⁠. Step 2 shows $P(A|B) = 1$ ⁠. Never true.
(A) would need $A \subset B$ as well, that is $A = B$ ⁠; the condition only gives $B \subset A$ ⁠.

$P(E) = \frac{n(E)}{n(S)}$ on a fair die. Basics: classical probability

Result

The condition gives $P(A \cap B) = P(B)$ ⁠, so $P(A|B) = 1$ ⁠: option (B). (Textbook answer: B.)

Check it with numbers

In the die example $P(A|B) = 1$ and $P(B|A) = \frac13 \approx 0.3333$ ⁠. The verify script takes every pair of events on a die with $P(A)$ and $P(B)$ not zero that satisfies the condition: 665 pairs. (B) holds in all of them, (A) only in the 63 pairs with $A = B$ ⁠, and (C) and (D) in none.

Common mistakes

Choosing (A). The condition puts $B$ inside $A$ ⁠, not $A$ inside $B$ ⁠; so it is $A$ that is certain given $B$ ⁠.
Reading "P(A and B)" as $P(A) + P(B)$ ⁠. "And" is the intersection $A \cap B$ ⁠.
Choosing (C) because " $B$ adds nothing". $B$ adds nothing outside $A$ ⁠, but $B$ itself can still happen.

What to take away

$P(A) + P(B) - P(A \cap B)$ is $P(A \cup B)$ ⁠; " $P(A \cup B) = P(A)$ " means $B$ lies inside $A$ ⁠.
$B \subset A$ goes with $P(A|B) = 1$ (compare Question 1 (i) and Question 11).

Every fact used, and where it comes from

Fact	Used in	Source
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	Step 1	Class XI (Basics: the addition rule)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \neq 0$	Steps 2, 4	Section 13.2, Definition 1 (Lesson: conditional probability)
$P(B) = P(A \cap B) + P(A' \cap B)$	Step 3	Class XI axioms (Basics: Venn diagrams)
$P(E) = \frac{n(E)}{n(S)}$ on a fair die	Step 4	Classical probability, Class XI (Basics: classical probability)

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Miscellaneous · Q13 (MCQ): P(A)+P(B)−P(A∩B)=P(A)P(A) + P(B) - P(A \cap B) = P(A)P(A)+P(B)−P(A∩B)=P(A) gives P(A∣B)=1P(A\mid B) = 1P(A∣B)=1⁠, option (B)

Basics you need

Solution

Simplify the condition

Apply the definition of P(A∣B)P(A|B)P(A∣B)

What it means for the sets

Examine every option

Every fact used, and where it comes from

Miscellaneous · Q13 (MCQ): $P(A) + P(B) - P(A \cap B) = P(A)$ gives $P(A\mid B) = 1$ ⁠, option (B)

Apply the definition of $P(A|B)$