Chapter 13 · Probability

Miscellaneous · Q11 (MCQ): $P(B\mid A) = 1$ gives $A \subset B$ ⁠, option (A)

Miscellaneous ExerciseConditional probability · Section 13.2

The question

Choose the correct answer.

If A and B are two events such that $P(A) \neq 0$ and $P(B | A) = 1$ ⁠, then

(A) $A \subset B$ (B) $B \subset A$ (C) $B = \phi$ (D) $A = \phi$

Exam answer

$P(B|A) = \frac{P(A \cap B)}{P(A)} = 1$ ⁠, so $P(A \cap B) = P(A)$ ⁠.

Then $P(A \cap B') = P(A) - P(A \cap B) = 0$ ⁠, so (with equally likely outcomes) $A \cap B' = \phi$ ⁠: every outcome of $A$ is in $B$ ⁠.

Answer: (A) $A \subset B$

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

The one idea this problem is built on

$P(B|A) = 1$ says: once we know $A$ has happened, $B$ is certain. That happens when every outcome of $A$ is also an outcome of $B$ ⁠, that is, when $A$ lies inside $B$ ⁠. It is the reverse of Question 1 (i), which showed that $A \subset B$ gives $P(B|A) = 1$ ⁠.

Given, to find, and the plan

Given: Events $A$ and $B$ of the same sample space, with $P(A) \neq 0$ and $P(B|A) = 1$ ⁠.
To find: Which of the four statements must be true.
Plan: (1) Unpack $P(B|A) = 1$ with the definition. (2) Turn the result into a statement about sets. (3) Test every option on a die.

Basics you need

1Conditional probability (Section 13.2)

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

2Splitting an event

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

3Subsets

$A \subset B$ means every outcome of $A$ is an outcome of $B$ ⁠; equivalently $A \cap B' = \phi$ ⁠. Basics: events as sets

Solution

Unpack $P(B|A) = 1$

Since $P(A) \neq 0$ ⁠, the definition gives
$\begin{aligned} &\frac{P(A \cap B)}{P(A)} = 1, \\ &\text{so } P(A \cap B) = P(A). \end{aligned}$
Multiplying both sides by $P(A)$ is allowed because $P(A) \neq 0$ ⁠. The overlap of $A$ with $B$ has the same probability as all of $A$ ⁠.

Section 13.2, Definition 1. Lesson: conditional probability
From probabilities to sets

Split $A$ into the part inside $B$ and the part outside $B$ ⁠:
$\begin{aligned} P(A \cap B') &= P(A) - P(A \cap B) \\ &= P(A) - P(A) = 0. \end{aligned}$
With equally likely outcomes, as in this chapter, an event of probability 0 has no outcomes. So $A \cap B' = \phi$ ⁠: no outcome of $A$ lies outside $B$ ⁠, which means $A \subset B$ ⁠.

$P(B|A) = 1$ ⁠: given $A$ ⁠, the outcome is certainly in $B$ ⁠, so $A$ sits inside $B$ ⁠.

"Probability 0" becomes "empty" because every outcome of a finite sample space with equally likely outcomes has a positive probability $\frac{1}{n(S)}$ ⁠. (In more general sample spaces, where single outcomes can have probability 0, $P(B|A) = 1$ only says that the part of $A$ outside $B$ has probability 0; option (A) is the statement intended here.)

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

Throw a fair die, $S = \{1, 2, 3, 4, 5, 6\}$ ⁠, and take $A = \{2\}$ ⁠, $B = \{2, 4, 6\}$ ⁠. Then $P(A) = \frac16 \neq 0$ and $P(B|A) = \frac16 \div \frac16 = 1$ ⁠, so this pair satisfies the question.
- (A) $A \subset B$ ⁠. True here, and Step 2 proves it always holds. Correct.
- (B) $B \subset A$ ⁠. False here: 4 is in $B$ but not in $A$ ⁠. Not always true.
- (C) $B = \phi$ ⁠. If $B = \phi$ ⁠, then $P(A \cap B) = 0$ and $P(B|A) = 0$ ⁠, not 1. Never true.
- (D) $A = \phi$ ⁠. Then $P(A) = 0$ ⁠, which the question rules out. Never true.
An option is correct only if it holds for every pair of events that satisfies the conditions; one counterexample rules an option out.

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

Result

$P(B|A) = 1$ gives $P(A \cap B') = 0$ ⁠, so $A \subset B$ ⁠: option (A). (Textbook answer: A.)

Check it with numbers

The verify script takes every pair of events $A$ ⁠, $B$ on a fair die with $P(A) \neq 0$ ⁠: 4032 pairs. $P(B|A) = 1$ in exactly 665 of them, and in every one of these $A \subset B$ ⁠. Among the 665, $B \subset A$ holds only in the 63 pairs with $A = B$ ⁠, and $B = \phi$ in none.

Common mistakes

Choosing (B). $P(B|A)$ is about $B$ inside the world of $A$ ⁠; it is $A$ that must lie in $B$ ⁠.
Thinking $P(B|A) = 1$ means $B$ is certain. Only given $A$ ⁠; on its own $P(B)$ can be less than 1 (here $\frac12$ ⁠).
Choosing (D). $A = \phi$ would make $P(A) = 0$ ⁠, and then $P(B|A)$ is not even defined.

What to take away

$P(B|A) = 1$ and $A \subset B$ go together (for equally likely outcomes): this question and Question 1 (i) are the two directions.
To turn a probability statement into a set statement, split an event into disjoint pieces.

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	Section 13.2, Definition 1 (Lesson: conditional probability)
$P(A) = P(A \cap B) + P(A \cap B')$	Step 2	Class XI axioms (Basics: Venn diagrams)
$A \cap B' = \phi$ means $A \subset B$	Step 2	Sets, Class XI (Basics: events as sets)
$P(E) = \frac{n(E)}{n(S)}$ on a fair die	Step 3	Classical probability, Class XI (Basics: classical probability)

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Miscellaneous · Q11 (MCQ): P(B∣A)=1P(B\mid A) = 1P(B∣A)=1 gives A⊂BA \subset BA⊂B⁠, option (A)

Basics you need

Solution

Unpack P(B∣A)=1P(B|A) = 1P(B∣A)=1

From probabilities to sets

Examine every option

Every fact used, and where it comes from

Miscellaneous · Q11 (MCQ): $P(B\mid A) = 1$ gives $A \subset B$ ⁠, option (A)

Unpack $P(B|A) = 1$