Conditional probability

Lesson overview

Statistics — advanced probability

**Conditional probability** is the probability of an event given that another event has already occurred. We write $P(A|B)$ to mean 'the probability of $A$ given $B$ '. $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ **Tree diagrams** are especially useful for conditional probability: the probabilities on the second set of branches depend on what happened on the first branch. For dependent events (e.g. picking items without replacement): - The probabilities on the second branch change based on the first outcome - $P(A \text{ and } B) = P(A) \times P(B|A)$

Conditional probability with a tree diagram

A bag contains $3$ red and $2$ blue counters. One is picked and NOT replaced. A second is picked. Find the probability both are red.

First red $\dfrac{3}{5}$

3 red out of 5

Second red $\dfrac{2}{4}$

2 red out of 4 left

Both red $\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{3}{10}$

Without replacement, the denominator changes on the second pick.

Using the conditional probability formula

If $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cap B) = 0.2$ , find $P(A|B)$ .

Divide the intersection by the given event's probability.

Venn diagrams for conditional probability

In a Venn diagram, $P(A|B)$ means we restrict our attention to just the $B$ circle and ask: what proportion of $B$ is also in $A$ ?

P(A|B)

$\dfrac{P(A \cap B)}{P(B)}$

given B, what's the chance of A?

P(B|A)

$\dfrac{P(A \cap B)}{P(A)}$

given A, what's the chance of B?

The condition tells you the denominator.

Worked example

In a group of $20$ students, $12$ play football, $8$ play rugby, and $3$ play both. A student who plays rugby is chosen at random. What is the probability they also play football?

We need $P(\text{football} | \text{rugby})$ .
Using the formula: $P(F|R) = \dfrac{P(F \cap R)}{P(R)}$ .
$P(F \cap R) = \dfrac{3}{20}$ and $P(R) = \dfrac{8}{20}$ .
$P(F|R) = \dfrac{3/20}{8/20} = \dfrac{3}{8}$ .

Try it

Identify which event is the condition (the 'given' part), then use it as the denominator.

Question 1

A bag has $4$ green and $6$ yellow counters. Two are picked without replacement. What is $P(\text{second is green} | \text{first is green})$ ?

Question 2

If $P(A \cap B) = 0.15$ and $P(B) = 0.3$ , what is $P(A|B)$ ?

Question 3

Two cards are drawn from a standard pack without replacement. What is $P(\text{both aces})$ ?

Question 4

If $P(A|B) = P(A)$ , what does this tell us about events $A$ and $B$ ?

Question 5

In a class, $P(\text{passes maths}) = 0.7$ and $P(\text{passes maths and science}) = 0.42$ . What is $P(\text{passes science} | \text{passes maths})$ ?

Common mistakes

Watch for these when working through the lesson.

Using $P(A)$ as the denominator instead of $P(B)$ when finding $P(A|B)$ .
Forgetting that without replacement changes the denominator on the second branch.
Assuming events are independent when they are actually dependent (e.g. picking without replacement).

Conditional probability

Lesson overview

Worked example

Try it

Common mistakes

Related topics

Independent practice will plug in here