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Probability Topics: Two Basic Rules of Probability

Module by: Dr. Barbara Illowsky, Susan Dean

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

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.

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 1

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 2

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 3

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 4

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 1

2815028150

Problem 2

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

Solution 2

8015080150

Problem 3

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

Solution 3

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 4

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 5

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 1

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 2

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 3

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 4

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 5

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 6

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 one of the following is true: (1). P ( A 2 B ) = P ( A ) ;P( A 2 B)=P(A); (2) P ( B 2 A ) = P ( B ) ;P( B 2 A)=P(B); (3) 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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