Conditional probability

What you will be able to do

• shrink the sample space when you are told that an event has happened;
• use $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and read it from a count, a table or a diagram;
• tell $P(A \mid B)$ apart from $P(B \mid A)$.

The die behind the screen

You roll a fair die and hide it behind a screen. The chance of a six is $\dfrac{1}{6}$. Now a friend peeks and says, "The number is even." Is the chance of a six still $\dfrac{1}{6}$?

No. The number can only be $2$, $4$ or $6$, and these three are equally likely. So the chance of a six is $\dfrac{1}{3}$. The news made every odd number impossible. The sample space shrank from six outcomes to three. That is the whole idea of this lesson.

The definition

Let $A$ and $B$ be two events of the same experiment, with $P(B) > 0$. The conditional probability of $A$ given $B$ is

Read the bar as the word "given". The event $B$ is the new sample space. On top is the part of $B$ where $A$ also happens. At the bottom is all of $B$. For the die, $B$ is "even", $A$ is "six", and $A \cap B$ is "six", so $P(A \mid B) = \dfrac{1/6}{3/6} = \dfrac{1}{3}$.

A word on notation. Many Indian books write $P(A/B)$. A-Level, IB and AP write $P(A \mid B)$. A slash looks like a division, so this book keeps the bar. All of them mean the same thing.

When all outcomes are equally likely, the formula becomes a count:

Count the outcomes in $B$. Then count how many of those are also in $A$.

A worked example with two dice

Two fair dice are thrown. Given that the sum is $8$, what is the chance that at least one die shows a six?

1. The new sample space. A sum of $8$ happens in five ways: $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$ and $(6,2)$.
2. The part with a six. Two of these ways have a six: $(2,6)$ and $(6,2)$.
3. The ratio. The probability is $\dfrac{2}{5}$.

Now check with the formula. The sum is $8$ in $5$ of the $36$ outcomes, and the sum is $8$ with a six in $2$ of them. So $P(A \mid B) = \dfrac{2/36}{5/36} = \dfrac{2}{5}$. The counting method and the formula always agree.

Reading a two-way table

A class has $60$ students. Of them, $36$ study music, $20$ are in the football team and $12$ do both.

• $P(\text{football} \mid \text{music}) = \dfrac{12}{36} = \dfrac{1}{3}$. Look only at the $36$ music students.
• $P(\text{music} \mid \text{football}) = \dfrac{12}{20} = \dfrac{3}{5}$. Look only at the $20$ footballers.

The two answers differ because the group after the bar is different. In general $P(A \mid B)$ and $P(B \mid A)$ are different numbers.

Three facts about conditional probability

Once $B$ is given, probabilities behave as they always do:

• $0 \le P(A \mid B) \le 1$, and $P(B \mid B) = 1$;
• $P(A' \mid B) = 1 - P(A \mid B)$, where $A'$ is the complement of $A$ (written $\bar{A}$ or $A^c$ in some books);
• $P(A \cup C \mid B) = P(A \mid B) + P(C \mid B) - P(A \cap C \mid B)$.

Finding a conditional probability from other probabilities

Suppose $P(A) = 0.6$, $P(B) = 0.4$ and $P(A \cup B) = 0.7$. First find the overlap: $P(A \cap B) = 0.6 + 0.4 - 0.7 = 0.3$. Then

The numerator is the same in both. Only the denominators differ.

Practise with new numbers

Try it in Python

The program lists all $36$ outcomes of two dice and finds a conditional probability by counting. It uses exact fractions, so the answers are written as fractions and not as decimals.

Common mistakes

• Dividing by the wrong probability. $P(A \mid B)$ divides by $P(B)$, the event after the bar.
• Swapping the two events. "$A$ given $B$" and "$B$ given $A$" are different questions with different answers.
• Counting outcomes outside $B$. After "given $B$", ignore every outcome that is not in $B$.
• Using an impossible condition. $P(A \mid B)$ is not defined when $P(B) = 0$.

Remember these

• Given $B$, the sample space is $B$, and $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ with $P(B) > 0$.
• For equally likely outcomes, $P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$.
• $P(A' \mid B) = 1 - P(A \mid B)$.
• $P(A \mid B)$ and $P(B \mid A)$ share a numerator but have different denominators.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A fair die is thrown. Given that the number is greater than $2$, what is the probability that it is even?

2. For two events, $P(A \cap B) = 0.18$ and $P(B) = 0.45$. Find $P(A \mid B)$.

3. Of $80$ cyclists, $50$ have a light, $30$ have a bell and $20$ have both. Find the probability that a cyclist has a light, given that the cyclist has a bell. Then find the probability of a bell, given a light.

4. Two fair coins are tossed. Given that at least one is a head, what is the probability that both are heads?

5. If $P(A \mid B) = 0.3$, find $P(A' \mid B)$.

References

• OpenStax, Introductory Statistics, Chapter 3, sections on the two basic rules of probability and on contingency tables.
• NCERT, Mathematics Part II for Class XII, Chapter 13, section on conditional probability.