Embedded Files
Probability Rules
Probability Rules
In this section, we will go over some key rules of probability. Specifically, we will cover the multiplication rule, the addition rule, and the conditional rule.
The Multiplication Rule
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
(The probability of A given B equals the probability of A and B divided by the probability of B.)
It is possible that two events are independent. This means that the probability of one event occurring is not impacted by the probability of the other event occurring. For example, the probability of getting a heads when tossing a fair coin AND the probability of getting a number greater than 5 when rolling a fair dice are independent events. Whether or not you get a heads does not impact whether or not you will get a number greater than 5 when you roll the dice. When events are independent, the probability rules for multiplication become simpler. 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
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).
It is possible that two events are mutually exclusive. This means that the probability of both events occurring is 0. For example, the probability of getting a heads and a tails if you flip a fair coin one time is 0 because the two events are mutually exclusive. 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).
The Conditional Rule
The Conditional Rule
The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is
where P(B) is greater than zero.
For example, suppose we toss one fair, six-sided die. The sample space
S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S).
We get the same result by using the formula. Remember that S has six outcomes.
Watch this video for an example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome.
EXAMPLE 1
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. He likes New Zealand a little better, so 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.
PRACTICE 1
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?
What is the probability that Carlos makes either the first goal or the second goal?
Are A and B independent?
Are A and B mutually exclusive?
PRACTICE 2
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event that Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?
PRACTICE 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 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.
What is the probability that the member is a novice swimmer?
What is the probability that the member practices four times a week?
What is the probability that the member is an advanced swimmer and practices four times a week?
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?
Are being a novice swimmer and practicing four times a week independent events? Why or why not?
PRACTICE 4
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports.
What is the probability that a senior is taking a gap year?
PRACTICE 5
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
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?
PRACTICE 6a
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.
What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
Given that the woman has breast cancer, what is the probability that she tests negative?
What is the probability that the woman has breast cancer AND tests negative?
What is the probability that the woman has breast cancer or tests negative?
Are having breast cancer and testing negative independent events?
Are having breast cancer and testing negative mutually exclusive?
PRACTICE 6b
Refer to the information in Example 6a. Now assume 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).
PRACTICE 7
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′).
References:
References:
CC LICENSED CONTENT, SHARED PREVIOUSLY
OpenStax, Statistics, Two Basic Rules of Probability. Provided by: OpenStax. Located at: http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.34:21/Introductory_Statistics. License: CC BY: Attribution
Introductory Statistics . Authored by: Barbara Illowski, Susan Dean. Provided by: Open Stax. Located at: http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44. License: CC BY: Attribution. License Terms: Download for free at http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44
ALL RIGHTS RESERVED CONTENT
Ex: Probability of Events that are NOT Mutually Exclusive Events. Authored by: Mathispower4u. Located at: https://youtu.be/z-1VvourLsA. License: Public Domain: No Known Copyright
DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013).
Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013).
“Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).
“Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013).
Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013).
Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013).
Data from U.S. Census Bureau.
Data from the Wall Street Journal.
Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013).
Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013).
Answers
Answers
Practice 1.
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) = 0.585Carlos makes the first and second goals with probability 0.585.
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.
No, they are not, because P(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.4230.423 ≠ 0.585 = P(B AND A)So, P(B AND A) is not equal to P(B)P(A).
No, they are not because P(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero.
Practice 2.
P(D and C) = P(D|C)P(C) = (0.85)(0.75) = 0.6375. This is the probability that Helen makes the first and second free throws.
Practice 3.
28/150
80/150
40/150
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. No, these are not independent events.
P(novice AND practices four times per week) = 0.0667P(novice)P(practices four times per week) = 0.09960.0667 ≠ 0.0996
Practice 4.
(200-140-40)/200 = 20/200 = 0.1
Practice 5.
0.1625
0.6875
No
No
Practice 6a.
P(B) = 0.143; P(N) = 0.85
P(N|B) = 0.02
P(B AND N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029
P(B OR N) = P(B) + P(N) – P(B AND N) = 0.143 + 0.85 – 0.0029 = 0.9901
No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N).
No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero.
Practice 6b.
0.98
0.1401
0.857
0.15
Practice 7.
P(B′) = 0.60
P(D AND B) = P(D|B)P(B) = 0.20
P(B|D)=P(B AND D)P(D)=0.20/0.30=0.66
P(D AND B′) = P(D) – P(D AND B) = 0.30 – 0.20 = 0.10
P(D|B′) = P(D AND B′)P(B′) = (P(D) – P(D AND B))(0.60) = (0.10)(0.60) = 0.06
Report abuse