4.11 Formal conditional probability and independence (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

The formal definition of conditional probability

The probability of $A$ given that $B$ has happened is $P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \qquad P(B)\ne0.$ Knowing that $B$ has happened reduces the sample space to $B$ . We then ask what fraction of $B$ is also in $A$ .

Order matters: $P(A\mid B)$ and $P(B\mid A)$ are usually different. With $P(A)=0.4$ , $P(B)=0.5$ and $P(A\cap B)=0.12$ , we get $P(A\mid B)=0.24$ but $P(B\mid A)=0.3$ .

Key termsconditional probabilityreduced sample space

Common mistake

Dividing by the wrong event. In $P(A\mid B)$ you divide by $P(B)$ , the event after the bar.

Exam tip

Translate words first: 'given that', 'if' and 'of those who' all introduce the condition after the bar.

Section 2

The multiplication form and combining branches

Rearranging the definition gives the multiplication rule $P(A\cap B)=P(B)P(A\mid B)=P(A)P(B\mid A).$ This is how you work along the branches of a tree diagram (described in words at this level). If it rains with probability 0.35 and Kofi is late with probability 0.6 when it rains, then $P(R\cap L)=0.35\times0.6=0.21$ .

To find the overall probability of an event, add the probabilities of the separate routes to it: $P(L)=P(R)P(L\mid R)+P(R')P(L\mid R')=0.21+0.13=0.34.$ You can then reverse the condition with the definition: $P(R\mid L)=\frac{0.21}{0.34}=0.618$ .

Key termsmultiplication rule

Common mistake

Using $P(L\mid R)=0.6$ as if it were $P(R\mid L)$ . Reversing a condition always needs the definition.

Section 3

Working with complements

Many questions need $P(A\mid B')$ . Use $P(A\cap B')=P(A)-P(A\cap B), \qquad P(B')=1-P(B),$ so $P(A\mid B')=\frac{P(A)-P(A\cap B)}{1-P(B)}.$ With $P(A)=0.4$ , $P(B)=0.5$ and $P(A\cap B)=0.12$ : $P(A\mid B')=\frac{0.28}{0.5}=0.56$ .

Also, $P(A'\mid B)=1-P(A\mid B)$ , because given $B$ , either $A$ happens or it does not.

Key termscomplement

Common mistake

Writing $P(A\mid B')=1-P(A\mid B)$ . That is wrong: the complement goes on $A$ , not on the condition. $1-P(A\mid B)=P(A'\mid B)$ .

Section 4

Independent events and testing for independence

$A$ and $B$ are independent if knowing that one has happened does not change the probability of the other: $P(A\mid B)=P(A)=P(A\mid B').$ This is equivalent to $P(A\cap B)=P(A)P(B).$ To test for independence, calculate one side of any of these and compare numerically. Conclude 'independent' only if they are equal, and always state the numbers you compared. For example, $P(A)P(B)=0.2\ne0.12=P(A\cap B)$ , so $A$ and $B$ are not independent.

Do not confuse this with mutually exclusive events, for which $P(A\cap B)=0$ . Two events with non-zero probabilities that are mutually exclusive cannot be independent, because if one happens the other cannot.

Key termsindependent eventsmutually exclusive

Common mistake

Assuming events are independent in order to test whether they are independent. Use the given data to find $P(A\cap B)$ separately.

Exam tip

With $P(A\cup B)$ given, independence lets you substitute $P(A\cap B)=P(A)P(B)$ and solve for an unknown.

Section 5

Conditional probability from counts

When data are given as counts, find probabilities by dividing by the size of the condition group. In a survey of 200 students where 120 travel by bus and 40 of those play tennis, $P(T\mid B)=\frac{40}{120}=\frac{1}{3}.$ For independence you would need $n(T\cap B)=\frac{n(T)\,n(B)}{n(U)}=\frac{80\times120}{200}=48$ . Interpret your conclusion in context, e.g. 'bus users are less likely to play tennis'.

Key termstwo-way table

Exam tip

For $P(T\mid B)$ , the denominator is the number in $B$ , not the total of 200.

Must know

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\cap B)=P(B)P(A\mid B)$ .
$P(A\mid B)\ne P(B\mid A)$ in general.
$P(A\cap B')=P(A)-P(A\cap B)$ .
Independent events: $P(A\mid B)=P(A)=P(A\mid B')$ , equivalently $P(A\cap B)=P(A)P(B)$ .
Test independence with a numerical comparison, and state the conclusion.
Mutually exclusive ( $P(A\cap B)=0$ ) is not the same as independent.

That's the notes covered.

Carry on to the next subtopic.