9.3 Two Basic Rules of Probability
When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.
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) = [latex]\frac{\text{P(A AND B)}}{\text{P(B)}}[/latex]
(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).
EXAMPLE 9.14
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.
EXAMPLE 9.15
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.
Problem
a. What is the probability that he makes both goals?
b. What is the probability that Carlos makes either the first goal or the second goal?
c. Are A and B independent?
d. Are A and B mutually exclusive?
Solution
a. 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.
b. 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.
c. 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).
d. No, they are not because P(A and B) = 0.585.
To be mutually exclusive, P(A AND B) must equal zero.
TRY IT 9.15
EXAMPLE 9.16
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 four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.
a. What is the probability that the member is a novice swimmer?
b. What is the probability that the member practices four times a week?
c. What is the probability that the member is an advanced swimmer and practices four times a week?
d. 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?
e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
Solution
a. [latex]\frac{28}{150}[/latex]
b. [latex]\frac{80}{150}[/latex]
c. [latex]\frac{40}{150}[/latex]
d. 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.
e. No, these are not independent events.
P(novice AND practices four times per week) = 0.0667
P(novice)P(practices four times per week) = 0.0996
0.0667 ≠ 0.0996
TRY IT 9.16
EXAMPLE 9.17
Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.
Let: M = math class, S = speech class, M|S = math given speech
Problem
- What is the probability that Felicity enrolls in math and speech?
Find P(M AND S) = P(M|S)P(S). - What is the probability that Felicity enrolls in math or speech classes?
Find P(M OR S) = P(M) + P(S) – P(M AND S). - Are M and S independent? Is P(M|S) = P(M)?
- Are M and S mutually exclusive? Is P(M AND S) = 0?
Solution
TRY IT 9.17
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
- Find P(B AND D).
- Find P(B OR D).
EXAMPLE 9.18
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 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.
Problem
a. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
b. Given that the woman has breast cancer, what is the probability that she tests negative?
c. What is the probability that the woman has breast cancer AND tests negative?
d. What is the probability that the woman has breast cancer or tests negative?
e. Are having breast cancer and testing negative independent events?
f. Are having breast cancer and testing negative mutually exclusive?
Solution
a. P(B) = 0.143; P(N) = 0.85
b. P(N|B) = 0.02
c. P(B AND N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029
d. P(B OR N) = P(B) + P(N) – P(B AND N) = 0.143 + 0.85 – 0.0029 = 0.9901
e. No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N).
f. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero.
TRY IT 9.18
EXAMPLE 9.19
Refer to the information in Example 9.18. P = tests positive.
- Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P|B) = 1 – P(N|B).
- What is the probability that a woman develops breast cancer and tests positive. Find P(B AND P) = P(P|B)P(B).
- What is the probability that a woman does not develop breast cancer. Find P(B′) = 1 – P(B).
- What is the probability that a woman tests positive for breast cancer. Find P(P) = 1 – P(N).
Solution
a. 0.98; b. 0.1401; c. 0.857; d. 0.15
TRY IT 9.19
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
- Find P(B′).
- Find P(D AND B).
- Find P(B|D).
- Find P(D AND B′).
- Find P(D|B′).
Attribution
By Barbara Illowsky and Susan Dean, Introductory Statistics, Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction. Jun 23, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License CC BY-NC-SA 4.0. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
