<< Chapter < Page Chapter >> Page >
This module introduces the multiplication and addition rules used when calculating probabilities.

The multiplication rule

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

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

(The probability of A given B equals the probability of A and B divided by the probability of B .)

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

The addition rule

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

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

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

  • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35 .
  • P(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 = 0.95 . Note that the probability that he does not choose to go anywhere on vacation must be 0.05 .

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.

A = the event Carlos is successful on his first attempt. P(A) = 0.65 . B = the event Carlos is successful on his second attempt. P(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.

What is the probability that he makes both goals?

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

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

Carlos makes the first and second goals with probability 0.585.

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

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

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

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

Are A and B independent?

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

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

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

Are A and B mutually exclusive?

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

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

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):

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

28 150

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

80 150

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

40 150

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?

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.

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

No, these are not independent events.

P(novice AND practices 4 times per week) = 0.0667
P(novice) P(practices 4 times per week) = 0.0996
0.0667 0.0996

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 negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.

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

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

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

P(N|B) = 0.02

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

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

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

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

Are having breast cancer and testing negative independent events?

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

Are having breast cancer and testing negative mutually exclusive?

No. P(B AND N) = 0.0029 . For B and N to be mutually exclusive, P(B AND N) must be 0.

Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Engr 2113 ece math. OpenStax CNX. Aug 27, 2010 Download for free at http://cnx.org/content/col11224/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Engr 2113 ece math' conversation and receive update notifications?

Ask