Conditional probability

A conditional probability is the probability of an event when another event is already known to have happened. The probability of $\displaystyle A$ given $\displaystyle B$ is written $\displaystyle P(A/B)$ (and also $\displaystyle P(A|B)$).

Knowing that $\displaystyle B$ happened narrows the sample space: only the outcomes of $\displaystyle B$ are left.

In the table of the 450 students from the last lesson:

You met a student from year 12. What is the probability that the student volunteers?

• The student is known to be in year 12, so the choice is only among the 140 students of year 12 – their row is now the whole sample space.
• Of them, 72 volunteer.

Of all the students $\displaystyle {\frac{2}{5}}={0.4}$ volunteer, and of the year-12 students – $\displaystyle \frac{18}{35}\approx0.51$.

In a table of numbers the cell is divided by the row’s total. Dividing both by the number of all the students, 450, leaves the fraction as it was – and this is the definition of conditional probability:

The definition needs $\displaystyle P(B)>0$. Multiplying by $\displaystyle P(B)$ gives the probability of an intersection from a conditional probability:

Often the data are percentages of a group, that is, conditional probabilities. Multiplying them by the size of the group gives the cells of the table.

In a high school 55% of the students are girls.
40% of the girls and 60% of the boys play in a sports team.
$\displaystyle G$ – a girl; $\displaystyle S$ – plays in a sports team.

• $\displaystyle {P(G\cap S)}={P(G)\cdot P(S/G)}={0.55\cdot0.4}={0.22}$;
• $\displaystyle {P(\overline{G}\cap S)}={0.45\cdot0.6}={0.27}$.

The area model draws the table: a square of area 1, a strip for the girls and a strip for the boys as wide as their probabilities, and each strip cut by the conditional probabilities. The area of each rectangle is a cell of the table, and the rectangles of $\displaystyle S$ are marked:

And among the team players, the probability of a girl:

$\displaystyle P(A/B)$ and $\displaystyle P(B/A)$ are two different questions: their numerator is the same intersection, but the denominator differs – once $\displaystyle P(B)$ and once $\displaystyle P(A)$.

• At the school: $\displaystyle {P(V/T)}={\frac{72}{140}}={\frac{18}{35}}$ – among the year-12 students, but $\displaystyle {P(T/V)}={\frac{72}{180}}={\frac{2}{5}}$ – among the volunteers.
• The probability that a person is taller than 1.90 metres given that he plays in the basketball league is close to 1. The probability that he plays in the league given that he is taller than 1.90 metres is close to 0: there are far more tall people than basketball players.

The difference is largest when one of the events is very rare and the other is common.

$\displaystyle P(A/B)$ says how much of $\displaystyle B$ is also inside $\displaystyle A$:

• $\displaystyle P(A/B)$ close to 0 – there is hardly an intersection: $\displaystyle A$ and $\displaystyle B$ are close to being disjoint.
• $\displaystyle P(A/B)$ close to $\displaystyle P(A)$ – knowing that $\displaystyle B$ happened hardly changes the chance of $\displaystyle A$.
• $\displaystyle P(A/B)$ close to 1 – almost all of $\displaystyle B$ is inside $\displaystyle A$. If $\displaystyle {P(A/B)}={1}$, then whenever $\displaystyle B$ happens $\displaystyle A$ happens too.

A clinic checked 1000 patients in the winter:

A patient is chosen at random.

The patient is known to be vaccinated. What is the probability that the patient had flu?

The patient is known not to be vaccinated. What is the probability that the patient had flu?

The patient is known to have had flu. What is the probability that the patient was vaccinated?

70% of the workers in a company come to work by car, and the rest another way.
20% of those who come by car are late at least once a week.
5% of those who come another way are late at least once a week.
Write the area of each rectangle in the area model, and mark the rectangles of the workers who are late.

What is the probability that a worker chosen at random is late at least once a week?

A worker is known to be late at least once a week. What is the probability that the worker comes by car?

$\displaystyle {P(A)}={0.5}$, $\displaystyle {P(B)}={0.4}$, $\displaystyle {P(A\cup B)}={0.7}$.

Work out $\displaystyle P(A\cap B)$.

Work out $\displaystyle P(A/B)$.

Work out $\displaystyle P(B/A)$.

$\displaystyle {P(B)}={0.6}$, $\displaystyle {P(A/B)}={0.25}$, $\displaystyle {P(A)}={0.3}$.

Work out $\displaystyle P(A\cap B)$.

Work out $\displaystyle P(B/A)$.

Work out $\displaystyle P(A/\overline{B})$.

Compare $\displaystyle P(A/B)$ and $\displaystyle P(A/\overline{B})$ with $\displaystyle P(A)$. What does knowing that $\displaystyle B$ happened do to the chance of $\displaystyle A$?