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Conditional Probability


Conditional probability is the likelihood that an event will occur given that another event has already occurred. There is a special notation for conditional probability. Suppose that A and B are two events and that the event A has already happened. Then the likelihood that the event B will also happen is said to be the probability of B conditional on the occurrence of A. The notation for this is  P(B | A), and may be read as the probability of B conditional on A, or as the probability of B given that A has already occurred.

What follows is a Venn diagram illustrating both events, where the intersection of A and B indicates the simultaneous occurrences of A and B.

In the case that the event A has already occurred, we focus only on A. This yields the following diagram.

The conditional probability  P(B | A) measures the likelihood that B will occur once A has occurred.

Of course, we can reverse the situation by assuming that the event B has already happened. Focusing only on the event B yields the following diagram.

Given that the event B has occurred, the likelihood that A will also occur is measured by the conditional probability  P(A | B).

Conditional probabilities are used to calculate the joint probability of two events through the Multiplication Rule.

P(A and B) = P(A) · P(B | A)

P(A and B) = P(B) · P(A | B).

Example

The RCMP in a small Alberta community know that 30% of home owners leave their doors unlocked. Crime records indicate that 5% of homes whose doors are left unlocked are burglarized.

  1. What probabilities are given in these statements? Describe them in terms of the relevant events.
  2. What percentage of homes in that community are likely to be left with the doors unlocked and be burglarized?

Solution to 1

Two probabilities, 30% or 0.3 and 5% or 0.05, are given. The relevant events are described below.

Let A be the event that a house is left with its doors unlocked.
Let B be the event that a house is burglarized.
Then 0.3 = P(A).
The 0.05 is the probability that a house is burglarized, given that its doors have been left unlocked. This is the conditional probability of the event B conditional on the prior occurrence of the event A.
Therefore, 0.05 = P(B | A).

Answer to 1

0.3 = P(A)  and  0.05 = P(B | A)

Solution to 2

By its wording, the question is essentially asking for the probability that a home in that community is both left with its doors unlocked and is burglarized. This is the joint probability of the events A and B which may be found using one or the other of the versions of the Multiplication Rule. We look at the probabilities given and choose the appropriate one.

P(A and B) = P(A) · P(B | A) = 0.3 · 0.05 = 0.015 .

Answer to 2

The percentage of homes in that community that are likely to be left with the doors unlocked and be burglarized is 1.5%.

Independent Events

  • The Simplest Case of Conditional Probability
  • The Simplest Case of the Multiplication Rule

Two events A and B are said to be independent if the occurence of one does not affect the occurence of the other. In other words, the probability that A occurs remains the same whether B has occurred or not. Moreover, the same is true in reverse. The probability that B occurs remains unaffected by the occurrence of A. The happenings of independent events are completely independent of one another.

The conditional probability of independent events simplifies to their original probability of happening.

P(A | B) = P(A)   and   P(B | A) = P(B)

The Multiplication Rule for independent events simplifies to:

P(A and B) = P(A)· P(B | A) = P(A)· P(B)

NOTE:  This does not mean that independent events cannot happen at the same time. In fact, independent events MUST be able to occur concurrently. Otherwise, if A and B are independent and mutually exclusive, then, by the Multiplication Rule,

P(A and B) = P(A)· P(B) = 0 .

This, in turn, implies that either P(A) = 0 or P(B) = 0 or both are 0, meaning that one or the other or both of the events cannot occur at all.

Therefore, events CANNOT be both mutally exclusive and independent!

Example

A local supermarket has two computers on its premises which are not networked. One is newer than the other. The probability that the newer one will break down in any given month is 0.04. The probability that the older one will break down in any given month is 0.07.

  1. Are these events independent?
  2. Are these events mutually exclusive?
  3. What is the probability that the two computers will both break down in any given month?
  4. What is the probability that either one or the other will break down in any given month?

Solution

Let A be the event that the newer computer breaks down in any given month.
Let B be the event that the older computer breaks down in any given month.
Then P(A) = 0.04  and  P(B) = 0.07.

Answer to 1:

A and B are independent events since either one of the computers is not affected by the breakdown of the other computer. They operate independently.

Answer to 2:

A and B are not mutually exclusive since there is a possibility that both computers could break down in any given month. Moreover, independent events cannot be mutually exclusive (see the argument above).

Answer to 3:

The probability that both computers will break down in any given month is obtained using the Multiplication Rule for independent events.

P(A and B) = P(A) · P(B) = 0.04 · 0.07 = 0.0028.

Answer to 4:

The probability that either one or the other will break down in any given month is obtained using the Addition Rule.

P(A or B) = P(A) + P(B) - P(A and B) = 0.04 + 0.07 - 0.0028 = 0.1072

 

Last Updated by SAS October 07, 2026, 16:00:54