Chapter 13 · Probability

Example 1: $P(A|B)$ when $P(B) = \frac{9}{13}$ and $P(A \cap B) = \frac{4}{13}$

ExamplesConditional probability · Section 13.2

The question

If $P(A) = \dfrac{7}{13}$ ⁠, $P(B) = \dfrac{9}{13}$ and $P(A \cap B) = \dfrac{4}{13}$ ⁠, evaluate $P(A|B)$ ⁠.

Exam answer

Given $P(B) = \frac{9}{13}$ and $P(A\cap B) = \frac{4}{13}$ ⁠.

By the definition of conditional probability,

\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \dfrac{\;\frac{4}{13}\;}{\frac{9}{13}} \\ &= \frac{4}{13}\times\frac{13}{9} \end{aligned}

$\therefore\ P(A|B) = \dfrac{4}{9}$ ⁠.

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

The one idea this problem is built on

"Given B" means B becomes the whole world. Once we know that $B$ has happened, only the outcomes inside $B$ still count, and $A$ can now happen only through the part it shares with $B$ ⁠. So the conditional probability is the share of $B$ taken up by $A \cap B$ ⁠:

P(A|B) = \frac{P(A \cap B)}{P(B)},

provided $P(B) \ne 0$ ⁠.

Given, to find, and the plan

Given: $P(A) = \frac{7}{13}$ ⁠, $P(B) = \frac{9}{13}$ ⁠, $P(A \cap B) = \frac{4}{13}$ ⁠.
To find: $P(A|B)$ ⁠, the probability of $A$ given that $B$ has occurred.
Plan: Write the definition; check that $P(B) \ne 0$ ⁠; substitute; divide the two fractions; then check that the data are possible and that the multiplication rule gives back $P(A \cap B)$ ⁠.

Basics you need

1Conditional probability

For events $E$ ⁠, $F$ of the same experiment with $P(F) \ne 0$ ⁠, $P(E|F) = \frac{P(E \cap F)}{P(F)}$ (Definition 1, Section 13.2). The event after the bar goes in the denominator. Lesson: conditional probability

2Dividing fractions

$\frac{a}{c} \div \frac{b}{c} = \frac{a}{c} \times \frac{c}{b} = \frac{a}{b}$ ⁠: with the same denominator, the denominators cancel. Basics: dividing by a fraction

3Two events cut S into four regions

"A only", "both", "B only" and "neither". Their probabilities add up to 1. Basics: Venn diagrams

Solution

Write the definition and check the condition
$P(A|B) = \frac{P(A \cap B)}{P(B)},$
which may be used because $P(B) = \frac{9}{13} \ne 0$ ⁠.

The question gives exactly the two numbers the definition needs: the probability of both events, and the probability of the event we are told has happened. The definition divides by $P(B)$ ⁠, so it only makes sense when $P(B) \ne 0$ ⁠; here $P(B) = \frac{9}{13}$ ⁠, so we may use it.

Definition 1, Section 13.2. Lesson: conditional probability
Substitute the given values
$P(A|B) = \dfrac{\;\frac{4}{13}\;}{\frac{9}{13}}.$
The top is $P(A \cap B) = \frac{4}{13}$ and the bottom is $P(B) = \frac{9}{13}$ ⁠. The value $P(A) = \frac{7}{13}$ is not needed for $P(A|B)$ ⁠; it is used only in the checks below.

Lesson: conditional probability
Divide the fractions
$\begin{aligned} P(A|B) &= \frac{4}{13} \times \frac{13}{9} \\ &= \frac{4}{9}. \end{aligned}$
Dividing by $\frac{9}{13}$ is the same as multiplying by its flip $\frac{13}{9}$ ⁠. The 13s cancel, so the answer is simply 4 parts out of 9.

Basics: dividing by a fraction
See it in a Venn diagram

Measure every probability in thirteenths. Then "A only" is $\frac{7}{13} - \frac{4}{13} = \frac{3}{13}$ ⁠, "B only" is $\frac{9}{13} - \frac{4}{13} = \frac{5}{13}$ ⁠, and "neither" is $1 - \frac{3 + 4 + 5}{13} = \frac{1}{13}$ ⁠.

The four regions, in thirteenths. Given $B$ ⁠, only circle $B$ counts: 9 thirteenths, of which 4 are in $A$ ⁠.

Inside circle $B$ there are $4 + 5 = 9$ thirteenths, and 4 of them lie in $A$ ⁠. That share, $\frac{4}{9}$ ⁠, is exactly what Step 3 found: knowing $B$ shrinks the world to circle $B$ ⁠.

Basics: Venn diagrams

Result

$P(A|B) = \dfrac{4}{9}$ ⁠.

(Textbook answer: $\frac{4}{9}$ ⁠.)

Check it with numbers

The data are possible. $P(A \cup B) = \frac{7}{13} + \frac{9}{13} - \frac{4}{13} = \frac{12}{13}$ ⁠, which is at most 1, and $P(A \cap B) = \frac{4}{13}$ is not larger than $P(A)$ or $P(B)$ ⁠.

The multiplication rule gives back the overlap. $P(B)\,P(A|B) = \frac{9}{13} \times \frac{4}{9} = \frac{4}{13}$ ⁠, which is the given $P(A \cap B)$ ⁠.

Decimals. $\frac{4}{9} \approx 0.4444$ ⁠, which lies between 0 and 1 as every probability must. It is smaller than $P(A) = \frac{7}{13} \approx 0.5385$ ⁠: knowing $B$ makes $A$ a little less likely, so $A$ and $B$ are not independent events (Section 13.4).

Common mistakes

Dividing by the wrong event. $P(A|B)$ divides by $P(B)$ ⁠, the event after the bar. Dividing by $P(A)$ gives $P(B|A) = \frac{4}{7}$ ⁠, a different number.
Multiplying instead of dividing. $\frac{4}{13} \times \frac{9}{13}$ is not a conditional probability.
Using $P(A)$ in the formula. The value $\frac{7}{13}$ is extra information for this question.

What to take away

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ ⁠: "both" over "the given one".
Check $P(B) \ne 0$ before dividing.
When both probabilities have the same denominator, the answer is just the ratio of the numerators.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Steps 1 to 3	Definition 1, Section 13.2 (Lesson: conditional probability)
$\frac{a}{c} \div \frac{b}{c} = \frac{a}{b}$	Step 3	Arithmetic of fractions (Basics: dividing by a fraction)
Four regions of a Venn diagram add to 1	Step 4	Class XI, Sets and Probability (Basics: Venn diagrams)
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	Check	Class XI, Probability (addition rule) (Basics: the addition rule)
$P(A \cap B) = P(B)\,P(A\|B)$	Check	Multiplication rule, Section 13.3 (Lesson: the multiplication rule)

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Example 1: P(A∣B)P(A|B)P(A∣B) when P(B)=913P(B) = \frac{9}{13}P(B)=139​ and P(A∩B)=413P(A \cap B) = \frac{4}{13}P(A∩B)=134​

Basics you need

Solution

Write the definition and check the condition

Substitute the given values

Divide the fractions

See it in a Venn diagram

Every fact used, and where it comes from

Example 1: $P(A|B)$ when $P(B) = \frac{9}{13}$ and $P(A \cap B) = \frac{4}{13}$