IB Maths AA 4.7 Notes

Bayes' theorem

Bayes' theorem is used to reverse conditional probabilities. It helps us find the probability of a cause given an observed result. In this course, it is used for a maximum of three events and links closely to conditional probability and independence.

Suppose events $A$ and $B$ satisfy $P(B)\gt 0$. Then Bayes' theorem states:

$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$

This formula is especially useful when $P(B\mid A)$ is known, but $P(A\mid B)$ is required.

Bayes' theorem lets us update probabilities using new information.

• $P(A)$ is the prior probability of $A$.
• $P(B\mid A)$ is the probability of observing $B$ if $A$ has occurred.
• $P(A\mid B)$ is the updated probability of $A$ after learning that $B$ has occurred.

From conditional probability, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ and $P(A\cap B)=P(B\mid A)P(A)$. Substituting gives Bayes' theorem, so it is a direct consequence of the conditional probability formula.

A bag contains $6$ red counters and $4$ blue counters. One counter is chosen at random. Let $A$ be the event that the counter is red, and let $B$ be the event that the counter is not blue. Find the probability that a red counter is chosen, given that a blue counter is not chosen.

Since 'not blue' means red in this case, $P(A\mid B)=1$. This simple example shows that once event $B$ is known, the probability of $A$ may change completely.

When there are two possible causes, the theorem changes slightly. Let's explore this through an example problem.

A school has $60\%$ day students and $40\%$ boarding students. Of the day students, $70\%$ play sport. Of the boarding students, $50\%$ play sport. A randomly chosen student is found to play sport. Find the probability that the student is a day student.

Let $D$ be the event that the student is a day student and $S$ be the event that the student plays sport.

We know:

• $P(D)=0.6$
• $P(S\mid D)=0.7$
• $P(S\mid D')=0.5$

First find $P(S)$ using total probability:

$P(S)=P(S\mid D)P(D)+P(S\mid D')P(D')=0.7(0.6)+0.5(0.4)=0.42+0.20=0.62$

Now apply Bayes' theorem:

$P(D\mid S)=\frac{P(S\mid D)P(D)}{P(S)}=\frac{0.7\times0.6}{0.62}=\frac{0.42}{0.62}\approx0.677$

So the probability that the student is a day student, given that they play sport, is about $0.677$.