International Baccalaureate · MYP 5

Conditional Probability

⌚ ~146 min ▢ 16 blocks ✓ 69 exercises ➣ Prerequisites : Probability

I Two-Way Tables and Conditional Frequencies

I.1 Conditional Frequencies

A data file describes individuals — people, animals, towns, cars — by several characteristics: species, region, profession, level of satisfaction. Crossing two of these characteristics in one table answers questions that a single list cannot: do the members of one group behave like those of another? which group holds most of a given category? A two-way table of counts is the first tool of data analysis, and reading it carefully is what keeps a sentence such as “most of the accidents with injuries happen in town” from being confused with “an accident in town is more often a serious one”.

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Definition — Two-Way Table of Counts

Two qualitative variables are observed on the same population of $N$ individuals. The two-way table of counts has one row for each category of the first variable and one column for each category of the second; the cell at row $A$ and column $B$ holds the count of the individuals that have both categories $A$ and $B$, written $n(A \cap B)$.

The last column holds the row totals: $n(A)$ is the number of individuals of category $A$.
The last row holds the column totals: $n(B)$ is the number of individuals of category $B$.
The bottom-right cell holds the grand total $N$, which is the sum of the row totals and the sum of the column totals.

Example

The $200$ members of a sports club are sorted by gender and by the sport they practise:

	Swimming	Athletics	Total
Women	$72$	$48$	$120$
Men	$32$	$48$	$80$
Total	$104$	$96$	$200$

Compute the marginal frequencies of the two variables.

Answer

The marginal counts are the totals of the rows and of the columns; each is divided by $N = 200$: $$\begin{aligned} f(\text{Women}) &= \frac{120}{200} = 0.6, & f(\text{Men}) &= \frac{80}{200} = 0.4, \\ f(\text{Swimming}) &= \frac{104}{200} = 0.52, & f(\text{Athletics}) &= \frac{96}{200} = 0.48. \end{aligned}$$ Check: $0.6 + 0.4 = 1$ and $0.52 + 0.48 = 1$. Women are $60\pourcent$ of the club, and swimming is the sport of $52\pourcent$ of the members.

Definition — Conditional Frequency

The conditional frequency of a category $B$ given a category $A$ observed at least once ($n(A) > 0$) is the proportion of the individuals of $A$ that also have $B$: $$\textcolor{colordef}{f_A(B) = \frac{n(A \cap B)}{n(A)}}.$$ It is computed inside the sub-population $A$: the numerator is the cell at row $A$ and column $B$, the denominator is the marginal count of $A$, not the grand total. The conditional frequencies given $A$ of all the categories of the other variable add up to $1$.

Method — Row Profiles and Column Profiles

To compare the groups defined by one variable:

Choose the variable that defines the groups: its categories are the rows (or the columns) of the table.
Divide each cell of a row by the total of that row (a row observed at least once): the row now reads as conditional frequencies, its row profile — the list of the $f_A(B)$ for the category $A$ of that row — which adds up to $1$.
Compare the profiles row by row: two groups with the same profile are distributed alike over the other variable.

The column profiles are obtained the same way, by dividing each cell by the total of its column. A row profile and a column profile answer two different questions and are not read one for the other.

Example

In the sports club of $200$ members (women $72$ swimmers and $48$ athletes, men $32$ swimmers and $48$ athletes):

Compute the frequency of swimming among the women and among the men. Do the two groups practise the same sports in the same proportions?
Compute the frequency of women among the swimmers and among the athletes.

Answer

The groups are the rows. Each cell is divided by its row total: $$\begin{aligned} f_{\text{Women}}(\text{Swimming}) &= \frac{n(\text{Women} \cap \text{Swimming})}{n(\text{Women})} = \frac{72}{120} = 0.6, \\ f_{\text{Men}}(\text{Swimming}) &= \frac{n(\text{Men} \cap \text{Swimming})}{n(\text{Men})} = \frac{32}{80} = 0.4. \end{aligned}$$ The row profiles are $(0.6\,;\,0.4)$ for the women and $(0.4\,;\,0.6)$ for the men: $60\pourcent$ of the women swim, against $40\pourcent$ of the men. The two groups do not practise the sports in the same proportions.
The groups are now the columns. Each cell is divided by its column total: $$\begin{aligned} f_{\text{Swimming}}(\text{Women}) &= \frac{72}{104} = \frac{9}{13} \approx 0.69, \\ f_{\text{Athletics}}(\text{Women}) &= \frac{48}{96} = 0.5. \end{aligned}$$ About $69\pourcent$ of the swimmers are women, and exactly half of the athletes. Note that $f_{\text{Women}}(\text{Swimming}) = 0.6$ and $f_{\text{Swimming}}(\text{Women}) \approx 0.69$ share the numerator $72$ but not the denominator: “$60\pourcent$ of the women swim” and “$69\pourcent$ of the swimmers are women” are two different facts.

Skills to practice

Computing Conditional Frequencies

II Conditional Probability

II.1 Probability Given an Event

Conditional probability updates a probability when new information becomes known. For instance, the probability of rain today rises once dark clouds appear: the extra information changes the probability.
Consider a bag of $5$ red, $3$ blue, and $2$ green balls ($10$ in total). The probability of picking a red ball is $\frac{5}{10} = \frac{1}{2}$. If we learn that the ball drawn is not blue, only the $5$ red and $2$ green balls ($7$ in total) remain possible, so the probability that it is red becomes $\frac{5}{7}$. This is conditional probability: the probability of an event (picking red) given that another event (the ball drawn is not blue) has occurred.
Formally, conditional probability is the likelihood of one event, say $F$, happening given that another event, $E$, has already taken place. We write it as $\PCond{F}{E}$, pronounced “the probability of $F$ given $E$”. It is a way to refine predictions with new information, and it is used everywhere—from weather forecasts to medical tests.

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Discover

Conditional probability can be explored with a two-way table showing the preferences of 100 students for maths, split by gender:

	Loves Maths	Does Not Love Maths	Total
Girls	35	16	51
Boys	30	19	49
Total	65	35	100

A student is picked at random, all students being equally likely. Some probabilities follow from the table:

Probability the student is a girl: $$\begin{aligned} P(\text{Girl}) &= \frac{\text{Number of girls}}{\text{Number of students}} \\ &= \frac{51}{100}. \end{aligned}$$
Probability the student loves maths and is a girl: $$\begin{aligned} P(\text{Loves Maths and Girl}) &= \frac{\text{Number of girls who love maths}}{\text{Number of students}} \\ &= \frac{35}{100}. \end{aligned}$$
Probability the student loves maths, given they are a girl:
Since the student is known to be a girl, only the 51 girls count: $$\begin{aligned} \PCond{\text{Loves Maths}}{\text{Girl}} &= \frac{\text{Number of girls who love maths}}{\text{Number of girls}} \\ &= \frac{35}{51}. \end{aligned}$$
Connecting to the formula:
Notice that: $$\begin{aligned} \PCond{\text{Loves Maths}}{\text{Girl}} &=\frac{35}{51}\\ &= \frac{35/100}{51/100}\\ &= \dfrac{P(\text{Loves Maths and Girl})}{P(\text{Girl})}. \end{aligned}$$ This pattern gives the general rule for conditional probability.

Definition — Conditional Probability

The conditional probability of event $F$ given event $E$ is the probability of $F$ occurring, given that $E$ occurs. It’s denoted $\PCond{F}{E}$ and calculated as: $$\textcolor{colordef}{\PCond{F}{E} = \frac{P(E \cap F)}{P(E)}}, \quad \text{where } P(E) > 0.$$

Two Notations

There are two notations for the conditional probability of $F$ given $E$. French textbooks write $P_E(F)$, with the known event $E$ as an index; international textbooks write $\PCond{F}{E}$, with the known event after a bar. They name the same number: $$P_E(F) = \PCond{F}{E} = \frac{P(E \cap F)}{P(E)}.$$ In both, $F$ is the event whose probability is computed and $E$ is the information already known. In the two-way table of the $100$ students, $35$ of the $51$ girls love maths: $P_{\text{Girl}}(\text{Loves Maths}) = \PCond{\text{Loves Maths}}{\text{Girl}} = \frac{35}{51}$.

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Example

A fair six-sided die has odd faces (1, 3, 5) painted green and even faces (2, 4, 6) painted blue. You roll it and see the top face is blue. What’s the probability it’s a 6?

Answer

Sample space: $\{1, 2, 3, 4, 5, 6\}$, 6 equally likely outcomes.
Event $E$ (face is blue): $\{2, 4, 6\}$, so $P(E) = \frac{3}{6}$.
Event $F$ (roll a 6): $\{6\}$.
Intersection $E \cap F$: $\{6\}$, so $P(E \cap F) = \frac{1}{6}$.
Conditional probability: $$\begin{aligned} \PCond{F}{E} &= \frac{P(E \cap F)}{P(E)}\\ &= \frac{\frac{1}{6}}{\frac{3}{6}}\\ &= \frac{1}{6} \times \frac{6}{3}\\ &= \frac{1}{3}. \end{aligned}$$
The probability of rolling a 6, given the face is blue, is $\frac{1}{3}$.

Proposition — Conditional Probability and Counts

An individual is chosen at random, all individuals being equally likely, in a population $U$ of $\Card{U}$ individuals. For two events $A$ and $B$, with $B$ containing at least one individual: $$\textcolor{colorprop}{P(A) = \frac{\Card{A}}{\Card{U}}} \qquad \text{and} \qquad \textcolor{colorprop}{\PCond{A}{B} = \frac{\Card{A \cap B}}{\Card{B}}}.$$ Here $\Card{A}$ is the number of elements of $A$. The conditional probability of $A$ given $B$ is the proportion of the individuals of $B$ that are also in $A$: the known event $B$ becomes the new population.

Proof

By the Definition “Equally Likely”, $P(A) = \dfrac{\Card{A}}{\Card{U}}$ and $P(B) = \dfrac{\Card{B}}{\Card{U}}$. Then: $$\begin{aligned} \PCond{A}{B} &= \frac{P(A \cap B)}{P(B)} && \text{(Definition ``Conditional Probability'')} \\ &= \frac{\Card{A \cap B}/\Card{U}}{\Card{B}/\Card{U}} && \text{(Definition ``Equally Likely'')} \\ &= \frac{\Card{A \cap B}}{\Card{U}} \times \frac{\Card{U}}{\Card{B}} && \text{(dividing by a fraction)} \\ &= \frac{\Card{A \cap B}}{\Card{B}}. \end{aligned}$$

Proposition — Conditioning Changes the Probability, Not the Events

Let $E$ be an event with $P(E) > 0$. Taking each event $F$ of the sample space $U$ to the number $\PCond{F}{E}$ defines a new probability on the same sample space: $\PCond{F}{E} \geq 0$ for every event $F$; $\PCond{U}{E} = 1$; and if $F \cap G = \emptyset$, then $\PCond{F \cup G}{E} = \PCond{F}{E} + \PCond{G}{E}$. The events are the same subsets of $U$ as before; only the numbers attached to them change. The new probability gives $1$ to $E$ and $0$ to $E'$, although $E'$ is not empty when $E \neq U$. Every rule proved for a probability therefore holds for it: for instance $\PCond{F'}{E} = 1 - \PCond{F}{E}$, which is why the probabilities on the branches leaving one node of a tree add up to $1$.

Proof

The three properties are the axioms of a probability, checked for the numbers $\PCond{F}{E}$.

Each $\PCond{F}{E} = \dfrac{P(E \cap F)}{P(E)}$ is the quotient of $P(E \cap F) \geq 0$ (Axiom 1) by $P(E) > 0$, so $\PCond{F}{E} \geq 0$.
$E \cap U = E$, so $\PCond{U}{E} = \dfrac{P(E \cap U)}{P(E)} = \dfrac{P(E)}{P(E)} = 1$. In the same way, $E \cap E = E$ gives $\PCond{E}{E} = 1$.
Let $F$ and $G$ be events with $F \cap G = \emptyset$. The events $E \cap F$ and $E \cap G$ have no outcome in common, since a common outcome would lie in $F \cap G$: they are mutually exclusive. Their union is $E \cap (F \cup G)$: an outcome of $E$ lies in $F \cup G$ exactly when it lies in $F$ or in $G$. So $$\begin{aligned}[t] \PCond{F \cup G}{E} &= \frac{P\big(E \cap (F \cup G)\big)}{P(E)} && \text{(Definition ``Conditional Probability'')}\\ &= \frac{P\big((E \cap F) \cup (E \cap G)\big)}{P(E)} && \text{(distributivity of $\cap$ over $\cup$)}\\ &= \frac{P(E \cap F) + P(E \cap G)}{P(E)} && \text{(Axiom 3)}\\ &= \PCond{F}{E} + \PCond{G}{E}. \end{aligned}$$
Finally $E \cap E' = \emptyset$, and $P(\emptyset) = 1 - P(U) = 0$ by the Proposition “Complement Rule” applied to $U$, whose complement is $\emptyset$. So $\PCond{E'}{E} = \dfrac{P(E \cap E')}{P(E)} = \dfrac{P(\emptyset)}{P(E)} = 0$, while $E'$ is not empty as soon as $E \neq U$.

The rule $F \mapsto \PCond{F}{E}$ satisfies the three axioms, so every rule deduced from the axioms alone holds for it; the Complement Rule gives $\PCond{F'}{E} = 1 - \PCond{F}{E}$.

Method — Reading a Conditional Probability in a Two-Way Table

To read $\PCond{A}{B}$ in a two-way table of counts:

Find the known event $B$ — the event after “given that” — and its row (or column).
Read $\Card{B}$ in the Total cell of that row (or column).
Read $\Card{A \cap B}$ in the cell where the row (or column) of $B$ meets the column (or row) of $A$.
Divide: $\PCond{A}{B} = \dfrac{\Card{A \cap B}}{\Card{B}}$.

The denominator is the total of the known event, read in the margin — not the grand total of the table.

Example

The $600$ pupils of a school are sorted by age and by lunch arrangement:

	Eats at school	Eats elsewhere	Total
Under 16	$180$	$60$	$240$
16 or over	$210$	$150$	$360$
Total	$390$	$210$	$600$

A pupil is chosen at random. Let $S$ be “the pupil is under 16” and $D$ be “the pupil eats at school”. Compute $P(S)$, $\PCond{D}{S}$ and $\PCond{S}{D}$, and state in words what the last two mean.

Answer

The pupils are equally likely. From the table: $$\begin{aligned} P(S) &= \frac{\Card{S}}{\Card{U}} = \frac{240}{600} = \frac{2}{5}, \\ \PCond{D}{S} &= \frac{\Card{S \cap D}}{\Card{S}} = \frac{180}{240} = \frac{3}{4}, \\ \PCond{S}{D} &= \frac{\Card{S \cap D}}{\Card{D}} = \frac{180}{390} = \frac{6}{13}. \end{aligned}$$ $\PCond{D}{S} = \frac{3}{4}$: among the pupils under 16, three out of four eat at school. $\PCond{S}{D} = \frac{6}{13}$: among the pupils who eat at school, six out of thirteen are under 16. The two probabilities share the numerator $180$ but not the denominator, and they answer two different questions.

Inverting the Conditioning

$\PCond{A}{B}$ and $\PCond{B}{A}$ are not equal in general: they share the numerator $\Card{A \cap B}$, but one divides by $\Card{B}$ and the other by $\Card{A}$. Everyday sentences often slide from one to the other. Take the bicycle rides made in a town during a year, and let $D$ be “the ride is made in daylight” and $I$ be “the cyclist is injured during the ride”. “Most of the injured cyclists were riding in daylight” is about $\PCond{D}{I}$, which would be large even if daylight were no more dangerous than night, simply because most rides are made in daylight; by itself it says nothing about $\PCond{I}{D}$, the risk of a daytime ride. A two-way table of counts keeps the two apart: the known event chooses the row or the column, hence the denominator.

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Example

In a population, $1\pourcent$ of the people have a certain illness. A screening test is positive for $99\pourcent$ of the ill people (its sensitivity) and negative for $95\pourcent$ of the healthy people (its specificity). A person is chosen at random; let $M$ be “the person is ill” and $T$ be “the test is positive”.

Fill in a two-way table of counts for a population of $10\,000$ people.
Compute $\PCond{T}{M}$ and $\PCond{M}{T}$, and compare them.

Answer

Out of $10\,000$ people, $1\pourcent$ are ill: $100$ ill and $9\,900$ healthy. The test is positive for $99\pourcent$ of the $100$ ill people, that is $99$ true positives, and negative for the other $1$, a false negative. It is negative for $95\pourcent$ of the $9\,900$ healthy people, that is $9\,405$, and positive for the other $495$.

Test positive $T$ Test negative $T'$ Total

Ill $M$ $99$ $1$ $100$

Healthy $M'$ $495$ $9\,405$ $9\,900$

Total $594$ $9\,406$ $10\,000$
The known event chooses the denominator: $$\begin{aligned} \PCond{T}{M} &= \frac{\Card{M \cap T}}{\Card{M}} = \frac{99}{100}, \\ \PCond{M}{T} &= \frac{\Card{M \cap T}}{\Card{T}} = \frac{99}{594} = \frac{1}{6}. \end{aligned}$$ Among the $100$ ill people, $99$ test positive; but among the $594$ people who test positive, only $99$ are ill, one out of six, because the $495$ false positives outnumber the $99$ true positives: the illness is rare, so the healthy people, even with a small error rate, supply most of the positive results.

Sometimes we know $P(E)$ and $\PCond{F}{E}$ and need the chance both $E$ and $F$ happen together—like finding the probability that a student is a girl who loves maths. This probability is called the joint probability $P(E \cap F)$.

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Proposition — Joint Probability Formula

For $P(E) > 0$ and $P(F) > 0$ respectively: $$P(E \cap F) = P(E) \times \PCond{F}{E}, \quad P(E \cap F) = P(F) \times \PCond{E}{F}.$$

Proof

By definition, for $P(E) > 0$, $\PCond{F}{E} = \dfrac{P(E \cap F)}{P(E)}$; multiplying both sides by $P(E)$ gives $P(E \cap F) = P(E) \times \PCond{F}{E}$. The second identity is the same statement with the roles of $E$ and $F$ exchanged, for $P(F) > 0$.

Skills to practice

Translating between Words and Symbols
Exploring Probabilities with Two-Way Tables
Calculating Conditional Probabilities
Calculating Conditional Probabilities in Real-World Problems
Reading and Inverting a Conditional Probability
Building a Two-Way Table from a Text

III Tree Diagrams

III.1 Independent Events and Probability Trees

Independent events are events where knowing that one event has happened does not change the probability that the other event happens. For example, when rolling two fair dice at the same time, the result of the first die does not change the chances for the second die — they are independent events.

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Definition — Independent Events

Two events $A$ and $B$ are independent when the probability that both happen (that is, $P(A\cap B)$ or $P(A \text{ and } B)$) is the product of their individual probabilities. This is called the multiplication rule for independent events: $$P(A \text{ and } B) = P(A) \times P(B)$$

Example

An experiment consists of the following two independent actions:

Tossing a fair coin.
Rolling a fair six-sided die.

What is the probability of getting tails and rolling a number greater than 4?

Answer

Let $T$ be the event “getting tails” and $N$ be the event “rolling a number greater than 4”.

The events are independent, so we can use the multiplication rule.
The probability of getting tails is $P(T) = \dfrac{1}{2}$.
The outcomes for a number greater than 4 are $\{5, 6\}$. There are 2 favourable outcomes out of 6 total outcomes. So, $P(N) = \dfrac{2}{6} = \dfrac{1}{3}$.
Now, we multiply the probabilities to find the probability of both events happening: $$\begin{aligned} P(T \text{ and } N) &= P(T) \times P(N) \\ &= \dfrac{1}{2} \times \dfrac{1}{3} \\ &= \dfrac{1}{6} \end{aligned}$$

So, the probability of getting tails and rolling a number greater than 4 is $\dfrac{1}{6}$.

Method — Using a Probability Tree Diagram

Draw branches for each step: Draw branches for the first event (coin toss) and then, from the end of each of those branches, draw the branches for the second event (die roll).
Write probabilities on each branch: The probabilities on the branches from a single point must add up to 1. Because the events are independent, the probabilities on the die-roll branches are the same after “Tails” and after “Heads”.
Multiply along the path: To find the probability of a combined event, multiply the probabilities along the path from start to finish. $$\textcolor{colorprop}{P(\text{``Tails'' and ``Number > 4''})=\frac{1}{2}\times \frac{1}{3}}$$

Skills to practice

Calculating Probabilities from a Tree Diagram

III.2 Conditional Probability Tree Diagrams

Example

A fair coin is tossed, then a fair die is rolled; the two trials are independent. Present the experiment in a table of probabilities and compute the probability of “Tails and an even number”.

Answer

Each row is a result of the coin, each column a result of the die; the margins carry $P(\text{Tails}) = P(\text{Heads}) = \frac{1}{2}$ and $\frac{1}{6}$ for each face. By independence, the probability of a cell is the product of its row total and its column total: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ for every cell, and the twelve cells add up to $1$.

	$1$	$2$	$3$	$4$	$5$	$6$	Total
Tails	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{2}$
Heads	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{2}$
Total	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$1$

“Tails and an even number” is the union of the three cells (Tails, $2$), (Tails, $4$), (Tails, $6$), which are mutually exclusive, so its probability is $3 \times \frac{1}{12} = \frac{1}{4}$. The multiplication rule for independent events gives the same value, $P(\text{Tails}) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$: the events “Tails” and “even” are independent too. A table cross-classifies the two trials; a tree follows them in order. Both carry the same probabilities.

From the Table to the Tree

The two-way table of the $100$ students can be read in two stages: first the gender, then, within the chosen gender, whether the pupil loves maths. Drawn as branches, these two stages make a tree: a first level with $P(\text{Girl}) = \frac{51}{100}$ and $P(\text{Boy}) = \frac{49}{100}$, then, along the “Girl” branch, $\PCond{\text{Loves Maths}}{\text{Girl}} = \frac{35}{51}$ and $\PCond{\text{Does Not Love Maths}}{\text{Girl}} = \frac{16}{51}$. The second-level branches carry conditional probabilities: that is what makes it a conditional probability tree, and what the Definition “Conditional Probability Tree Diagram” sets down.

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Definition — Conditional Probability Tree Diagram

A conditional probability tree visually organizes probabilities for a sequence of events:

Each branch from a node shows either an unconditional probability (e.g. $P(E)$) or a conditional probability (e.g. $\PCond{F}{E}$).
Events are labelled at the end of each branch.
The probabilities on the branches leaving one node add up to $1$ (sum rule): for instance $\PCond{F'}{E} = 1 - \PCond{F}{E}$.
The probability of the intersection of the events met along a path is the product of the probabilities along that path (product rule).

Example

The probability Sam coaches a game is $\frac{6}{10}$, and the probability Alex coaches is $\frac{4}{10}$. If Sam coaches, the probability that a randomly selected player is a goalkeeper is $\frac{1}{2}$; if Alex coaches, it is $\frac{2}{3}$.
Draw the tree diagram.

Answer

Define the events:
- $S$: Sam coaches.
- $G$: Player is goalkeeper.
Define the probabilities:
- $P(S) = \frac{6}{10}$ and $P(S') =1 - P(S)= \frac{4}{10}$.
- $\PCond{G}{S} = \frac{1}{2}$ and $\PCond{G'}{S} = 1 - \PCond{G}{S} = \frac{1}{2}$.
- $\PCond{G}{S'} = \frac{2}{3}$ and $\PCond{G'}{S'} = 1-\PCond{G}{S'}= \frac{1}{3}$.
Tree diagram:

Method — Finding $P(E \cap F)$ in a Tree

Identify the path where $E$ and $F$ both occur.
Multiply the probabilities along that path.

$$P(E \cap F) = \textcolor{colorprop}{P(E) \PCond{F}{E}}.$$

Example

For this probability tree,

find $P(S \cap G)$.

Answer

Path: $S$ to $G$ (highlighted):
Calculate: $$\begin{aligned} P(S \cap G) &= P(S) \times \PCond{G}{S}\\ &= \frac{6}{10} \times \frac{1}{2}\\ &= \frac{3}{10}. \end{aligned}$$

Skills to practice

Identifying Conditional Probability Tree Diagrams
Calculating Joint Probabilities with Trees
Drawing Conditional Probability Tree Diagrams

IV Independence and Conditional Probability

IV.1 Independence Seen through Conditional Probability

Two events are independent when the probability of both is the product of their probabilities. The words used to explain it, “knowing one does not change the chances of the other”, can now be made exact: they are a statement about a conditional probability. Independent events must not be confused with mutually exclusive events: two mutually exclusive events of non-zero probability are never independent, since $E \cap F = \emptyset$ gives $P(E \cap F) = P(\emptyset) = 0$, while $P(E) \times P(F) \neq 0$.

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Proposition — Conditional Characterisation of Independence

Let $E$ and $F$ be two events with $P(E) > 0$ and $P(F) > 0$. Then $$E \text{ and } F \text{ are independent} \iff \PCond{F}{E} = P(F) \iff \PCond{E}{F} = P(E).$$ Knowing that $E$ has happened does not change the probability of $F$, and knowing $F$ does not change the probability of $E$. The product form is the definition; the conditional form is what it means, and the one used when an event is given as known.

Proof

By definition of the conditional probability, $\PCond{F}{E} = \dfrac{P(E \cap F)}{P(E)}$. Since $P(E) > 0$, multiplying both sides of an equality by $P(E)$ is reversible: $$\PCond{F}{E} = P(F) \iff \frac{P(E \cap F)}{P(E)} = P(F) \iff P(E \cap F) = P(E) \times P(F).$$ The last equality is the definition of independence. It is symmetric in $E$ and $F$, so, dividing by $P(F) > 0$, it is also equivalent to $\dfrac{P(E \cap F)}{P(F)} = P(E)$, that is $\PCond{E}{F} = P(E)$.

Example

The $400$ pupils of a high school are sorted by lunch arrangement and by whether they play a musical instrument.

	Plays an instrument	Does not play	Total
Eats at school	$60$	$180$	$240$
Eats at home	$40$	$120$	$160$
Total	$100$	$300$	$400$

A pupil is chosen at random. Let $S$ be “the pupil eats at school” and $M$ be “the pupil plays an instrument”. Are $S$ and $M$ independent?

Answer

With the conditional form: $P(M) = \dfrac{100}{400} = \dfrac{1}{4}$ and, among the $240$ pupils who eat at school, $60$ play, so $\PCond{M}{S} = \dfrac{60}{240} = \dfrac{1}{4}$. Since $\PCond{M}{S} = P(M)$, the events are independent: knowing that the pupil eats at school does not change the chance that they play.
With the product: $P(S) \times P(M) = \dfrac{240}{400} \times \dfrac{100}{400} = \dfrac{3}{5} \times \dfrac{1}{4} = \dfrac{3}{20}$ and $P(S \cap M) = \dfrac{60}{400} = \dfrac{3}{20}$. Same conclusion.

Proposition — Independence and Complementary Events

If $E$ and $F$ are independent, then so are $E$ and $F'$, $E'$ and $F$, and $E'$ and $F'$: each of these pairs satisfies the product rule.

Proof

The events $E \cap F$ and $E \cap F'$ are mutually exclusive and their union is $E$. Hence $$\begin{aligned}[t] P(E \cap F') &= P(E) - P(E \cap F) && \scriptscriptstyle\text{(mutually exclusive pieces of } E\text{)} \\ &= P(E) - P(E)P(F) && \scriptscriptstyle\text{(independence)} \\ &= P(E)\bigl(1 - P(F)\bigr) \\ &= P(E)P(F') && \scriptscriptstyle\text{(complement rule)} \end{aligned}$$ So $E$ and $F'$ are independent. Exchanging the roles of $E$ and $F$ gives $E'$ and $F$; applying the result twice gives $E'$ and $F'$.

Skills to practice

Computing a Conditional Probability to Test Independence
Deciding whether Two Events are Independent

V Going further

Skills to practice

Inverting the Conditioning: False Positives
Questioning Intuitions about Chance
Investigating Independence in Two-Way Tables
Quiz

	Swimming	Athletics	Total
Women	\(72\)	\(48\)	\(120\)
Men	\(32\)	\(48\)	\(80\)
Total	\(104\)	\(96\)	\(200\)

	Eats at school	Eats elsewhere	Total
Under 16	\(180\)	\(60\)	\(240\)
16 or over	\(210\)	\(150\)	\(360\)
Total	\(390\)	\(210\)	\(600\)

	Test positive \(T\)	Test negative \(T'\)	Total
Ill \(M\)	\(99\)	\(1\)	\(100\)
Healthy \(M'\)	\(495\)	\(9\,405\)	\(9\,900\)
Total	\(594\)	\(9\,406\)	\(10\,000\)

	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	Total
Tails	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{2}\)
Heads	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{2}\)
Total	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(1\)

	Plays an instrument	Does not play	Total
Eats at school	\(60\)	\(180\)	\(240\)
Eats at home	\(40\)	\(120\)	\(160\)
Total	\(100\)	\(300\)	\(400\)