# 2.5 - Conditional Probability

2.5 - Conditional Probability

## Conditional Probability and Independence

The concept of conditional events and independent events determines whether or not one of the events has an effect on the probability of the other event. You can possibly imagine several daily conversations you may have that invoke these concepts. For instance, say you are discussing driving directions with a friend on the quickest way to get to some destination. The friend asks if leaving during rush hour would change these plans. If you respond that it does, this reflects the concept of "conditional" as the quickest way is conditioned, or affected, by the time of day your friend leaves. If you respond that it does not, this reflects the concept of "independence" as the quickest way is the same regardless of when your friend leaves. The time of day does not affect the quickest way.

Another example is practically any sporting event. For instance, a team might have a probability of 0.6 of winning the Super Bowl or a country a probability of 0.3 of winning the World Cup.

• Conditional Scenario: What if it rains the team's chances may change (for the better or possibly for the worse)? The probability of winning is affected by the weather - conditional
• Independent Scenario: What if the game is being played in an enclosed stadium? In such a case the weather may have no effect on the team's chances - independent.
Conditional Probability

The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol $\mid$ which is read as "given."

For example, $P(A\mid B)$ is read as "Probability of A given B."

Computing Conditional Probability

The Probability of A given B:

$P(A\mid B)=\dfrac{P(A \: \cap\: B)}{P(B)}$

The Probability of B given A:

$P(B\mid A)=\dfrac{P(B \: \cap\: A)}{P(A)}$

The Probability of the Intersection of Dependent Events

The probability of dependent events A and B derived from the formulas for conditional probability:

$P(A \cap B)=P(B) P(A|B)$

$P(B \cap A)=P(A) P(B|A)$

Note! Usually, $P(A|B) \neq P(B|A)$.
Marginal Probability
The probability of an event without reference to any other event or events occurring.

## Application

### Enrollment Figures

The two-way table below displays the World Campus enrollment from Fall 2015 in terms of level (undergraduate and graduate) and biological sex.

Female Male Total
Total 6027 6215 12242

Choose one student from the sample, what is the probability that the student is a female?

$P(F)=\dfrac{6027}{12242}$

If it is known that the student is a graduate student, what is the probability that the student is a female?

We now are interested in only graduate students who are female (2213). The total is just the number of graduate students (5000).

$P(F|G)=\dfrac{2213}{5000}$

Note that this is not the same as the probability that a selected student is a female graduate student (ie. $P(F \cap G$)).

In this scenario, the marginal probability is not the same as the conditional probability. This means that given the student is a graduate, changes the likelihood that the student is a female. This is reflective of events that are not independent i.e. they are considered dependent events. Independent events will be discussed in more detail later in the lesson.

## Example 2-6

Consider rolling two fair six-sided dice and the events C and D.

$C=\{\text{a 5 is rolled on the second die}\}$

$D=\{\text{the sum of the dice is 7}\}$,

find $P(C|D)$.

Let's approach this example in two ways: (1) using the sample space and (2) using the formulas above.

Approach 1:  The sample space

When we know that $D$ occurs, we only consider the pair of rolls in $D$ and forget all the other possibilities. Therefore, the conditional space becomes:

First Die
1 2 3 4 5 6

Second

Die

1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1
2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2
3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3
4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4
5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5
6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6

In set notation...

S={(1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Out of the six pairs that have a sum of 7, we can see only one pair, (2,5), has a 5 on the second die. Therefore,

$P(C|D)=\dfrac{1}{6}$

Approach 2:  From the previous example, we know $P(D)=\frac{6}{36}$, $P(D\cap C)=\frac{1}{36}$.  Therefore, using the formula above,

$P(C|D)=\dfrac{P(C\cap D)}{P(D)}=\dfrac{\frac{1}{36}}{\frac{6}{36}}=\dfrac{1}{6}$.

## Important Note Regarding Conditional Probabilities

For any two events, $A$ and $B$,
1. $P(A|B)=1-P(A^\prime|B)$
2. $P(A)=P(A\cap B)+P(A\cap B^\prime)=P(A|B)P(B)+P(A|B^\prime)P(B^\prime)$

The Venn diagrams below will help you visualize why this is true.

##### P(A∩B)
The first diagram shows $P(A\cap B)$.

$+$

##### $P(A\cap B^\prime)$
...when added to $P(A\cap B^\prime)$...

$=$

##### $P(A)$
...gives the $P(A)$.

  Link ↥ Has Tooltip/Popover Toggleable Visibility