Statistics Tutorial: Rules of Probability

Often, we want to compute the probability of an event from the known probabilities of other events. This lesson covers some important rules that simplify those computations.

Definitions and Notation

Before discussing the rules of probability, we state the following definitions:

.Two events are mutually exclusive if they have no sample points in common.
 .The probability that Event A occurs, given that Event B has occurred, is called a conditional probability.
 .The conditional probability of A, given B, is denoted by the symbol P(A|B).
 .The probability that event A will not occur is denoted by P(A').

Rule of Subtraction

The probability of a sample point ranges from 0 to 1.
The sum of probabilities of all the sample points in a sample space equals 1.

The rule of subtraction follows directly from these properties.

Rule of Subtraction The probability that event A will occur is equal to 1 minus the probability that event A will not occur.

P(A) = 1 - P(A')

Rule of Multiplication

The rule of multiplication applies to the following situation. We have two events from the same sample space, and we want to know the probability that both events occur.

Rule of Multiplication If events A and B come from the same sample space, the probability that both A and B occur is equal to the probability the event A occurs times the probability that B occurs, given that A has occurred.

P(A ∩ B) = P(A) * P(B|A)

Example 1

An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn without replacement from the urn. What is the probability that both of the marbles are black?

Solution: Let A = the event that the first marble is black; and let B = the event that the second marble is black.

We know the following:

In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore, P(A) = 4/10.
After the first selection, there are 9 marbles in the urn, 3 of which are black. Therefore, P(B|A) = 3/9.

Therefore, based on the rule of multiplication:
P(A ∩ B) = P(A) P(B|A)
P(A ∩ B) = (4/10)*(3/9) = 12/90 = 2/15


Example 2
Suppose we repeat the experiment of Example 1; but this time we select marbles with replacement. That is, we select one marble, note its color, and then replace it in the urn before making the second selection. When we select with replacement, what is the probability that both of the marbles are black?

Solution: Let A = the event that the first marble is black; and let B = the event that the second marble is black. We know the following:
In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore, P(A) = 4/10.
After the first selection, we replace the selected marble; so there are still 10 marbles in the urn, 4 of which are black. Therefore, P(B|A) = 4/10.

Therefore, based on the rule of multiplication:
P(A ∩ B) = P(A) P(B|A)
P(A ∩ B) = (4/10)*(4/10) = 16/100 = 4/25

Rule of Addition

The rule of addition applies to the following situation. We have two events from the same sample space, and we want to know the probability that either event occurs.

Rule of Addition If events A and B come from the same sample space, the probability that event A and/or event B occur is equal to the probability that event A occurs plus the probability that event B occurs minus the probability that both events A and B occur.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Note: Invoking the fact that P( A ∩ B ) = P( A )P( B | A ), the Addition Rule can also be expressed as
P(A ∪ B) = P(A) + P(B) - P(A) * P( B | A )

Example 1
A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30, , and (c) both fiction and non-fiction is 0.20. What is the probability that the student checks out a work of fiction, non-fiction, or both?

Solution: Let F = the event that the student checks out fiction; and let N = the event that the student checks out non-fiction. Then, based on the rule of addition:
P(F ∪ N) = P(F) + P(N) - P(F ∩ N)
P(F ∪ N) = 0.40 + 0.30 - 0.20 = 0.50

Example 2
A card is drawn randomly from a deck of ordinary playing cards. You win $10 if the card is a spade or an ace. What is the probability that you will win the game?

Solution: Let S = the event that the card is a spade; and let A = the event that the card is an ace. We know the following:
There are 52 cards in the deck.
There are 13 spades, so P(S) = 13/52.
There are 4 aces, so P(A) = 4/52.
There is 1 ace that is also a spade, so P(S ∩ A) = 1/52.

Therefore, based on the rule of addition:
P(S ∪ A) = P(S) + P(A) - P(S ∩ A)
P(S ∪ A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13