20 3.3 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 [latex]\displaystyle{P}{({A}{\mid}{B})}=\frac{{{P}{({A}\text{ AND } {B})}}}{{{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 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.
The probability that he chooses New Zealand is 0.6 and the probability that he chooses Alaska is 0.35.
Klaus can only afford to take one vacation. 

  1. What is the probability that he chooses either New Zealand or Alaska?
  2. What is the probability that he does not choose to go anywhere on vacation?
Show Answer
  1. Let A be New Zealand,  B be Alaska.
    P(Klaus chooses New Zealand) = P(A) = 0.6
    P(Klaus chooses Alaska ) = P(B) = 0.35
    P(Klaus chooses both New Zealand and Alaska) = P(A and B) = 0 as he can only afford one vacation.
    P(A OR B) = P(A) + P(B) – P(A and B) = 0.6 + 0.35 – 0 = 0.95.
    Therefore, the probability that he chooses either New Zealand or Alaska is 0.95.
  2. The probability that he does not choose to go anywhere on vacation
    = 1 – P( A or B)
    = 1 – 0.95
    = 0.05.

 

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

P(A) = 0.65, P(B) = 0.65, P(B|A) = 0.90.

  1. What is the probability that he makes both goals?
    Show Answer

    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.

  2. What is the probability that Carlos makes either the first goal or the second goal?
    Show Answer

    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.

  3. Are A and B independent?
    Show Answer

    P(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.423
    Since P(B AND A) [latex]\ne[/latex] P(B)*P(A), A and B are not independent.

  4. Are A and B mutually exclusive?
    Show Answer

    P(A and B) = 0.585.
    To be mutually exclusive, P(A AND B) must equal zero.
    Therefore, A and  are not mutually exclusive events.

 

Watch this video for another example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome.

Try It

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. The probability that Helen makes the first shot is 0.75. The probability that Helen makes the second shot is 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?

Show Answer

Let C be the event that Helen makes the first shot.
Let D be the event that Helen makes the second shot.
P(C) = 0.75, P(D) = 0.75, P(D|C) = 0.85
P(C AND D) = P(D AND C) = P(D|C)P(C) = (0.85)(0.75) = 0.6375
Helen makes the first and second free throws with probability 0.6375.

 

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 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.

  1. What is the probability that the member is a novice swimmer?
    Show Answer

    [latex]\frac{{28}}{{150}}[/latex]

  2. What is the probability that the member practices four times a week?
    Show Answer

    [latex]\frac{{80}}{{150}}[/latex]

  3. What is the probability that the member is an advanced swimmer and practices four times a week?
    Show Answer

    [latex]\frac{{40}}{{150}}[/latex]

  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?
    Show Answer

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

  5. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
    Show Answer

    No, these are not independent events.
    P(novice AND practices four times per week) = 0.0667
    P(novice) * P(practices four times per week) = 0.09960.
    P(novice AND practices four times per week)  [latex]\neq[/latex] P(novice) * P(practices four times per week)  

Try It

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?

Show Answer

The probability that a senior is taking a gap year = [latex]\frac{{{200}-{140}-{40}}}{{200}}=\frac{{20}}{{200}}={0.1}[/latex]

 

Example 4

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

  1. What is the probability that Felicity enrolls in math and speech?
  2. What is the probability that Felicity enrolls in math or speech classes?
  3. Are M and S independent?
  4. Are M and S mutually exclusive?

Solution

  1. Show Answer

    P(M AND S) = P(M|S)P(S) = 0.25 * 0.65 = 0.1625.

  2. Show Answer

    P(M OR S) = P(M) + P(S) – P(M AND S) = 0.2 + 0.65 – 0.1625 = 0.6875

  3. Show Answer

    P(M AND S) = 0.1625
    P(M) * P(S) = 0.2 * 0.65 = 0.13
    Since P(M AND S) [latex]\ne[/latex] P(M) * P(S), M and S are not independent.

  4. Show Answer

    Since P(M AND S) = 0.1625 [latex]\ne[/latex] 0, M and S are not mutually exclusie.

Try It

A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.

  1. Find P(B AND D).
    Show Answer

    P(B AND D) = P(D|B)P(B) = (0.5)(0.4) = 0.20.

  2. Find P(B OR D).
    Show Answer

    P(B OR D) = P(B) + P(D) − P(B AND D) = 0.40 + 0.30 − 0.20 = 0.50

 

Example 5

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.

  1. What is the probability that the woman develops breast cancer?
    Show Answer

    P(B) = 0.143

  2. What is the probability that woman tests negative?
    Show Answer

    P(N) = 0.85

  3. Given that the woman has breast cancer, what is the probability that she tests negative?
    Show Answer

    P(N|B) = 0.02

  4. What is the probability that the woman has breast cancer AND tests negative?
    Show Answer

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

  5. What is the probability that the woman has breast cancer or tests negative?
    Show Answer

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

  6. Are having breast cancer and testing negative independent events?
    Show Answer

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

  7. Are having breast cancer and testing negative mutually exclusive?
    Show Answer

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

Try It

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 going to college and plays sports?

Show Answer

Let A = student is a senior going to college. Let B = student plays sports.

Show Answer

P(A and B) = [latex]\frac{\text{number of seniors going to college and play sports}}{\text{total amount of seniors}}[/latex] = [latex]\frac{50}{200}[/latex] = [latex]\frac{1}{4}[/latex] 

 

Example 6

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, let N = tests negative, and let P = tests positive.
Suppose one woman is selected at random.

  1. Given that a woman develops breast cancer, what is the probability that she tests positive?
    Show Answer

    P(P|B) = 1 – P(N|B) = 1 – 0.02 = 0.98.

  2. What is the probability that a woman develops breast cancer and tests positive?
    Show Answer

    P(B AND P) = P(P|B)P(B) = 0.98 * 0.143 = 0.1401.

  3. What is the probability that a woman does not develop breast cancer?
    Show Answer

    P(B′) = 1 – P(B) = 1 – 0.143 = 0.857.

  4. What is the probability that a woman tests positive for breast cancer?
    Show Answer

    P(P) = 1 – P(N) = 1 – 0.85 = 0.15.

 

Try It

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.

  1. Find P(B′).
  2. Find P(D AND B).
  3. Find P(B|D).
  4. Find P(D AND B′).
  5. Find P(D|B′).
Show Answer
  1. P(B′) = 0.60
  2. P(D AND B) = P(D|B) * P(B) = 0.20
  3. P(B|D)=[latex]\frac{{{P}{({B}\text{ AND } {D})}}}{{{P}{({D})}}}=\frac{{{0.20}}}{{{0.30}}}={0.66}[/latex]
  4. P(D AND B′) = P(D) – P(D AND B) = 0.30 – 0.20 = 0.10
  5. P(D|B′) = P(D AND B′) * P(B′) = (P(D) – P(D AND B))(0.60) = (0.10)(0.60) = 0.06

 

References

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).

Concept Review

The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common.

Formula Review

The multiplication rule: P(A AND B) = P(A|B)P(B)

The addition rule: P(A OR B) = P(A) + P(B) – P(A AND B)

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