We are often interested in probabilities under certain conditions. For example what is the probability of it snowing tomorrow given the temperature today, or if an even number is rolled on a die what is the probability that it is a six. Giving conditions will reduce the sample space we are concerned with.

For two events $A$A and $B$B the probability of event $A$A occurring given that event $B$B has occurred has the notation $P\left(A|B\right)$P(A|B) which is read as the "probability of $A$A given $B$B".

Worked examples

Example 1

What is the probability of rolling a $6$6 given that an even number has been rolled?

Think: Let's define the two events involved, event $A$A: rolling a six and event $B$B: rolling an even number.

Do: Since we know an even number has been rolled the possible outcomes has reduced from $\left\{1,2,3,4,5,6\right\}${1,2,3,4,5,6} to $\left\{2,4,6\right\}${2,4,6}. Of these there is one favourable outcome which is both even and a six.

$P\left(A\|B\right)$`P`(`A`\|`B`)	$=$=	$\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes
	$=$=	$\frac{1}{3}$13

Reflect: If we consider how we found these figures we can come up with a rule for finding $P\left(A|B\right)$P(A|B). The number of outcomes possible was reduced to be the number of outcomes for event $B$B - since this was given to have happened. The number of favourable outcomes must be in the reduced set of outcomes and match event $A$A, hence it is the number of elements in $A\cap B$A∩B. We can think about this in terms of the number of outcomes for certain events, and we get the formula:

$P\left(A|B\right)=\frac{\text{number in }\left(A\cap B\right)}{\text{number in }\left(B\right)}$P(A|B)=number in (A∩B)number in (B)

Example 2

$56$56 students from two year 12 classes were surveyed and asked if they studied Mathematics and/or Physics. The results are shown in the Venn diagram below:

(a) If a student is selected at random what is the probability the student studies Mathematics?

This is not a conditional probability question and can be calculated using the number of students studying Mathematics and the total number of students.

$P\left(\text{Mathematics}\right)$`P`(Mathematics)	$=$=	$\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes
	$=$=	$\frac{35}{56}$3556
	$=$=	$\frac{5}{8}$58

(b) If a student is selected at random what is the probability that the student studies Mathematics given that they study Physics?

This is a conditional probability, we have been given that the student selected studies Physics. This will reduce the students we are interested in to just those studying Physics:

We now have a reduced total outcome of $25$25 students and of these$15$15 study Mathematics.

$P\left(\text{Mathematics\|Physics}\right)$`P`(Mathematics\|Physics)	$=$=	$\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes
	$=$=	$\frac{\text{number in }\left(\text{Mathematics}\cap\text{Physics}\right)}{\text{number in }\left(\text{Physics}\right)}$number in (Mathematics∩Physics)number in (Physics)
	$=$=	$\frac{15}{25}$1525
	$=$=	$\frac{3}{5}$35

(c) If a student is selected at random what is the probability that the student studies Physics given that they study Mathematics?

This is a conditional probability, we have been given that the student selected studies Mathematics. This will reduce the students we are interested in to just those studying Mathematics:

We now have a reduced total outcome of $35$35 students and of these $15$15 study Physics.

$P\left(\text{Physics\|Mathematics}\right)$`P`(Physics\|Mathematics)	$=$=	$\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes
	$=$=	$\frac{\text{number in }\left(\text{Physics}\cap\text{Mathematics}\right)}{\text{number in }\left(\text{Mathematics}\right)}$number in (Physics∩Mathematics)number in (Mathematics)
	$=$=	$\frac{15}{35}$1535
	$=$=	$\frac{3}{7}$37

Reflect: Notice in our example $P\left(A|B\right)\ne P\left(B|A\right)$P(A|B)≠P(B|A) and this is true in general. Can you see under what conditions they would be equal?

So we can calculate the probability of event A occurring given event B has occurred using:

$P\left(A|B\right)=\frac{\text{number in }\left(A\cap B\right)}{\text{number in }\left(B\right)}$P(A|B)=number in (A∩B)number in (B)

If we divide the top and bottom terms of this fraction by the total number of outcomes, each term would be expressible as a probability. This allows us to calculate conditional probabilities using the probabilities of events rather than the number of outcomes.

Conditional probability

For two events $A$A and $B$B, the probability of $A$A occurring given that$B$B has occurred is given by:

$P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$P(A|B)=P(A∩B)P(B)

If we know the conditional probability we can calculate the probability of $A$A and $B$B using either:

$P\left(A\cap B\right)=P\left(A|B\right)P\left(B\right)$P(A∩B)=P(A|B)P(B) OR $P\left(A\cap B\right)=P\left(B|A\right)P\left(A\right)$P(A∩B)=P(B|A)P(A)

Independent events using conditional probability

Recall that independent events are events where the occurrence of one event does not affect the probability of the other occurring. Such as rolling a dice and then tossing a coin, or tossing a coin repeatedly. Now in the context of conditional probability this would mean that the probability of event $A$A occurring given event $B$B has occurred should simply be the probability of $A$A - that is, it shouldn't matter if event $B$B occurred or not. Mathematically we can write this property as:

$P\left(A|B\right)=P\left(A\right)$P(A|B)=P(A)

From our conditional formula we see that for independent events:

$P\left(A\cap B\right)$`P`(`A`∩`B`)	$=$=	$P\left(A\|B\right)\times P\left(B\right)$`P`(`A`\|`B`)×`P`(`B`)
	$=$=	$P\left(A\right)\times P\left(B\right)$`P`(`A`)×`P`(`B`)

We can use this fact to test if two events are independent if we know the probability of $A$A, $B$B and the probability of (A and B).

Independent events

The following statements are true for any two independent events, $A$A and $B$B:

$P\left(A\cap B\right)=P\left(A\right)\times P\left(B\right)$P(A∩B)=P(A)×P(B)
$P\left(A|B\right)=P\left(A\right)$P(A|B)=P(A)
$P\left(B|A\right)=P\left(B\right)$P(B|A)=P(B)

Practice questions

QUESTION 1

A student creates the following diagram of their favourite animals.

How many $of the animals$ $have four legs$ ?
How many $of the animals$ $have four legs and stripes$ ?
What is the probability of selecting an animal with stripes given that it has four legs?

QUESTION 2

The probability of two independent events, $A$A and $B$B are, $P\left(A\right)=0.6$P(A)=0.6 and $P\left(B\right)=0.7$P(B)=0.7.

Determine the probability of:

Both $A$A and $B$B occurring.
Neither $A$A nor $B$B.
$A$A or $B$B or both.
$B$B but not $A$A.
$A$A given that $B$B occurs.

QUESTION 3

A flight departs from Melbourne to Sydney. The probability that the flight departs on time, given the weather is fine in Melbourne is $0.9$0.9, and the probability that the flight departs on time, given the weather is not fine in Melbourne is $0.7$0.7. The probability that the weather is fine on any particular day in July is $0.4$0.4.

By constructing a tree diagram or otherwise, find the probability that the flight from Melbourne to Sydney departs on time in a day in July.

Give your answer in its simplest form.
Find the probability that the weather is fine in Melbourne given that the flight departs on time on a day in July?

Give your answer in its simplest form.

8.02 Conditional probability