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Probability Topics: Two Basic Rules of Probability

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

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Summary: This module introduces the multiplication and addition rules used when calculating probabilities.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

The Multiplication Rule

If AA and BB are two events defined on a sample space, then: P(A AND B) = P(B)⋅P(A|B)P(A AND B)=P(B)⋅P(A|B).

This rule may also be written as : P(A|B)=P(A|B)= P(A AND B) P(B) P(A AND B) P(B)

(The probability of AA given BB equals the probability of AA and BB divided by the probability of BB.)

If A and B are independent, then P(A|B) = P(A)P(A|B) = P(A). Then P(A AND B) = P(A|B) P(B)P(A AND B) = P(A|B) P(B) becomes P(A AND B) = P(A) P(B)P(A AND B)=P(A) P(B).

The Addition Rule

If A A and BB are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B)P(A OR B)=P(A)+P(B)-P(A AND B).

If AA and BB are mutually exclusive, then P(A AND B) = 0P(A AND B)= 0. Then P(A OR B) = P(A) + P(B) - P(A AND B)P(A OR B)=P(A)+P(B)-P(A AND B) becomes P(A OR B) = P(A) + P(B)P(A OR B)= P(A)+P(B).

Example 1

Klaus is trying to choose where to go on vacation. His two choices are: AA = New Zealand and BB = Alaska

  • Klaus can only afford one vacation. The probability that he chooses AA is P(A) = 0.6P(A) = 0.6 and the probability that he chooses BB is P(B) = 0.35P(B)= 0.35.
  • P(A and B) = 0P(A and B) = 0 because Klaus can only afford to take one vacation
  • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95P(A OR B) =P(A) + P(B)=0.6+0.35 =0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.050.05.

Example 2

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game.

AA = the event Carlos is successful on his first attempt. P(A) = 0.65P(A)=0.65. BB = the event Carlos is successful on his second attempt. P(B) = 0.65P(B)=0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

Problem 1

What is the probability that he makes both goals?

Solution

The problem is asking you to find P(A AND B) = P(B AND A)P(A AND B)=P(B AND A). Since P(B|A) = 0.90P(B|A)=0.90:

P(B AND A) = P(B|A) P(A)  =  0.90 * 0.65 = 0.585 P(B AND A)= P(B|A) P(A) = 0.90*0.65=0.585 (1)

Carlos makes the first and second goals with probability 0.585.

Problem 2

What is the probability that Carlos makes either the first goal or the second goal?

Solution

The problem is asking you to find P(A OR B)P(A OR B).

P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715 P(A OR B)=P(A)+ P(B) -P(A AND B)=0.65+0.65-0.585=0.715 (2)

Carlos makes either the first goal or the second goal with probability 0.715.

Problem 3

Are AA and BB independent?

Solution

No, they are not, because P(B AND A)  =  0.585 P(B AND A) = 0.585.

P(B)  ⋅  P(A)  =  (0.65)  ⋅  (0.65)  =  0.423 P(B) ⋅ P(A) = (0.65) ⋅ (0.65) = 0.423(3)
0.423  ≠  0.585  =  P(B AND A) 0.423 ≠ 0.585 = P(B AND A)(4)

So, P(B AND A)P(B AND A) is not equal to P(B) ⋅ P(A) P(B) ⋅ P(A).

Problem 4

Are A A and BB mutually exclusive?

Solution

No, they are not because P(A and B)P(A and B) = 0.585.

To be mutually exclusive, P(A AND B)P(A AND B) must equal 0.

Example 3

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice 4 times a week. Thirty of the intermediate swimmers practice 4 times a week. Ten of the novice swimmers practice 4 times a week. Suppose one member of the swim team is randomly chosen. Answer the questions (Verify the answers):

Problem 1

What is the probability that the member is a novice swimmer?

Solution

2815028150

Problem 2

What is the probability that the member practices 4 times a week?

Solution

8015080150

Problem 3

What is the probability that the member is an advanced swimmer and practices 4 times a week?

Solution

4015040150

Problem 4

What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?

Solution

P(advanced AND intermediate)  =  0 P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

Problem 5

Are being a novice swimmer and practicing 4 times a week independent events? Why or why not?

Solution

No, these are not independent events.

P(novice AND practices 4 times per week) = 0.0667 P(novice AND practices 4 times per week)=0.0667 (5)
P(novice) ⋅ P(practices 4 times per week) = 0.0996 P(novice) ⋅ P(practices 4 times per week)=0.0996 (6)
0.0667 ≠ 0.0996 0.0667≠0.0996 (7)

Example 4

Studies show that, if she lives to be 90, about 1 woman in 7 (approximately 14.3%) will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is believed to be negative about 85% of the time. Let B B = woman develops breast cancer and let NN = tests negative.

Problem 1

What is the probability that a woman develops breast cancer? What is the probability that woman tests negative?

Solution

P(B) = 0.143P(B)= 0.143 ; P(N) = 0.85P(N)=0.85

Problem 2

Given that a woman has breast cancer, what is the probability that she tests negative?

Solution

P(N|B) = 0.02P(N|B)=0.02

Problem 3

What is the probability that a woman has breast cancer AND tests negative?

Solution

P(B AND N) = P(B) ⋅ P(N|B) = (0.143) ⋅ (0.02) = 0.0029P(B AND N)=P(B) ⋅ P(N|B)=(0.143) ⋅ (0.02)= 0.0029

Problem 4

What is the probability that a woman has breast cancer or tests negative?

Solution

P(B OR N) = P(B) + P(N) - P(B AND N) = 0.143 + 0.85 - 0.0029 = 0.9901P(B OR N)= P(B)+P(N) -P(B AND N)=0.143+0.85-0.0029=0.9901

Problem 5

Are having breast cancer and testing negative independent events?

Solution

No. P(N) = 0.85P(N)=0.85; P(N|B) = 0.02P(N|B)=0.02. So, P(N|B)P(N|B) does not equal P(N)P(N)

Problem 6

Are having breast cancer and testing negative mutually exclusive?

Solution

No. P(B AND N) = 0.0020P(B AND N)= 0.0020. For BB and NN to be mutually exclusive, P(B AND N)P(B AND N) must be 0.

Glossary

Independent Events:
The occurrence of one event has no effect on the probability of the occurrence of any other event. Events A and B are independent if any of the following is true:
  • P ( A | B ) = P ( A ) P(A|B)=P(A)
  • P ( B | A ) = P ( B ) P(B|A)=P(B)
  • P ( A and B ) = P ( A ) P ( B )P(AandB)=P(A)P(B)
.
Mutually Exclusive:
An observation cannot fall into more than one class (category). Being in one category prevents being in a mutually exclusive category.
Sample Space:
The set of all possible outcomes of an experiment.

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