Chapter 13 · Probability

Exercise 13.1 · Q16 (MCQ): $P(A|B)$ when $P(A) = \frac{1}{2}$ ⁠, $P(B) = 0$

Exercise 13.1Conditional probability · Section 13.2

The question

Choose the correct answer.

If $P(A) = \frac{1}{2}$ ⁠, $P(B) = 0$ ⁠, then $P(A|B)$ is

(A) 0 (B) $\frac{1}{2}$

Exam answer

By the definition of conditional probability, $P(A|B) = \frac{P(A\cap B)}{P(B)}$ only when $P(B)$ is non-zero.

Here $P(B) = 0$ ⁠, so the formula cannot be used.

Answer: (C) not defined

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

The one idea this problem is built on

$P(A|B)$ means "the chance of $A$ in the world where $B$ has happened". The definition $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ carries the condition $P(B) \ne 0$ ⁠. When $P(B) = 0$ there is no such world to shrink to, and the fraction would be a division by zero. So $P(A|B)$ is not defined: option (C). The value $P(A) = \frac12$ plays no part.

Given, to find, and the plan

Given: Events $A$ ⁠, $B$ with $P(A) = \frac12$ and $P(B) = 0$ ⁠.
To find: Which option describes $P(A|B)$ ⁠.
Plan: (1) Recall the definition with its condition. (2) Test the condition. (3) Examine every option.

Basics you need

1Conditional probability and its condition

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

2Division by zero

A fraction $\frac{a}{b}$ is the number that gives $a$ when multiplied by $b$ ⁠. With $b = 0$ there is no such single number, so $\frac{a}{0}$ is not defined. Basics: dividing by a fraction

3A smaller event has a smaller probability

$A\cap B$ is part of $B$ ⁠, so $0 \le P(A\cap B) \le P(B)$ ⁠. Basics: the rules of probability

Solution

Recall the definition, with its condition
$P(A|B) = \frac{P(A\cap B)}{P(B)},$
provided $P(B) \ne 0$ ⁠.

The book notes that this is valid only when $P(F) \ne 0$ ⁠, that is $F \ne \phi$ ⁠, and asks the reader why. The reason: conditioning on $B$ means using $B$ as the new sample space and measuring $A$ as a share of it. A share of something with probability 0 has no meaning.

Definition 1, Section 13.2. Lesson: conditional probability
Test the condition

Here $P(B) = 0$ ⁠, so the condition fails. Also $A\cap B$ is part of $B$ ⁠, so
$0 \le P(A\cap B) \le P(B) = 0,$
which forces $P(A\cap B) = 0$ ⁠. The formula would read $\frac{0}{0}$ ⁠.

$\frac{0}{0}$ has no single value: $0\times 1 = 0$ ⁠, $0\times\frac12 = 0$ and $0\times 7 = 0$ ⁠, so every number "works" and none is the answer. Hence $P(A|B)$ is not defined.

Division by zero is not defined. Basics: dividing by a fraction
Examine every option

(A) 0. The top $P(A\cap B)$ is 0, but the bottom is 0 too. A fraction is 0 only when its top is 0 and its bottom is not. Wrong.

(B) $\frac12$ ⁠. This is $P(A)$ ⁠. It would be the answer if knowing $B$ did not change the chance of $A$ ⁠, but that idea also needs $P(B) \ne 0$ ⁠. Here no value at all can be given. Wrong.

(C) not defined. The definition requires $P(B) \ne 0$ ⁠, and $P(B) = 0$ ⁠. Correct.

(D) 1. A conditional probability is 1 when $B$ is inside $A$ ⁠, as in $P(F|F) = 1$ (Property 1, Section 13.2.1), but that property also assumes $P(F) \ne 0$ ⁠. Wrong.

Options (A), (B) and (D) are all numbers, and each would need the definition to produce a value. The definition cannot be applied, so no number is correct.

Every property of conditional probability is stated for $P(F) \ne 0$ ⁠. Lesson: conditional probability

Result

$P(A|B)$ is not defined, because $P(B) = 0$ ⁠. The answer is (C).

(Textbook answer: C.)

Check it with numbers

Why no number can be chosen: roll a fair die and let $A = \{1, 2, 3\}$ ⁠, so $P(A) = 0.5$ ⁠. With $B_1 = \{1\}$ ⁠, $P(B_1) = \frac16 \approx 0.1667$ and $P(A|B_1) = 1$ ⁠; with $B_2 = \{4\}$ ⁠, also $P(B_2) \approx 0.1667$ ⁠, but $P(A|B_2) = 0$ ⁠. Small events can give any answer, and an event of probability 0, such as $B$ ⁠: "the die shows 7", gives nothing to divide by.

Common mistakes

Choosing (A) because the top is 0. The bottom is also 0; $\frac{0}{0}$ is not 0.
Choosing (B) because $B$ "cannot affect" $A$ ⁠. The value $P(A) = \frac12$ is a distraction: it does not enter the answer.
Forgetting the condition $P(F) \ne 0$ ⁠. It is part of the definition, not a footnote.

What to take away

$P(E|F)$ exists only when $P(F) \ne 0$ ⁠. Check this before using the formula.
Conditioning on an impossible event makes no sense: there is no world left to measure $E$ in.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$	Steps 1, 2	Section 13.2, Definition 1 and the note before it (Lesson: conditional probability)
$A\cap B \subseteq B$ ⁠, so $P(A\cap B) \le P(B)$	Step 2	Class XI, Probability (Basics: the rules of probability)
$P(F\|F) = 1$	Step 3, option (D)	Section 13.2.1, Property 1
Division by zero is not defined	Step 2	Arithmetic (Basics: dividing by a fraction)

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Exercise 13.1 · Q16 (MCQ): P(A∣B)P(A|B)P(A∣B) when P(A)=12P(A) = \frac{1}{2}P(A)=21​⁠, P(B)=0P(B) = 0P(B)=0

Basics you need

Solution

Recall the definition, with its condition

Test the condition

Examine every option

Every fact used, and where it comes from

Exercise 13.1 · Q16 (MCQ): $P(A|B)$ when $P(A) = \frac{1}{2}$ ⁠, $P(B) = 0$