2.2 Conditional Probability and the Multiplication Rule


  1. You can use probabilities from a two-way table to look for and describe relationships between two categorical variables. The following table displays information about cigarette smoking and diagnosis with hypertension for a group of patients at a medical clinic.
    Hypertension Diagnosis No Hypertension Diagnosis Total
    Smoker 48 24 72
    Nonsmoker 26 50 76
    Total 74 74 148
    1. If someone is a smoker, what is the probability they are diagnosed with hypertension?

      P(diagnosed with hypertension, given a smoker) = 48/72 = 0.6667 = 66.67%

    2. If someone is a nonsmoker, what is the probability they are diagnosed with hypertension?

      P(diagnosed with hypertension, given nonsmoker) = 26/76 = 0.3421 = 34.21%

    3. If someone is diagnosed with hypertension, what is the probability they are a smoker?

      P(smoker, given diagnosed with hypertension) = 48/74 = 0.6486 = 64.86%

    4. Given that someone does not have hypertension, what is the probability they are not a smoker?

      P(not a smoker, given do not have hypertension) = 50/74 = 25/37 = 0.6757 = 67.57%

  2. The following contingency table represents a bowl full of M&Ms and Skittles. Find the probabilities:
    Red Candy Yellow Candy Green Candy Blue Candy Orange Candy Brown or Purple Total Number
    M&Ms 109 102 142 183 187 105 828
    Skittles 202 188 145 0 154 156 845
    Total Number 311 290 287 183 341 261 1673
    1. One randomly selected a piece of candy:
      1. \(P(\mathrm{m} \& \mathrm{m} | \mathrm{Red})\)

        \(\frac{109}{311}=0.3505\)

      2. \(P(\operatorname{Red} | \mathrm{m} \& \mathrm{m})\)

        \(\frac{109}{828}=0.1316\)

      3. \(P(\text { Skittle } | \text { Yellow })\)

        \(\frac{188}{290}=0.6483\)

      4. \(P(\mathrm{Yellow} | \mathrm{Skittle})\)

        \(\frac{188}{845}=0.2225\)

    2. Two randomly selected pieces of candy:
      1. \(P(\text { Both Blue })\) with replacement

        \(\left(\frac{183}{1673}\right)\left(\frac{183}{1673}\right)=0.011965\) independent events

      2. \(P(\text { Both Blue })\) without replacement

        \(\left(\frac{183}{1673}\right)\left(\frac{182}{1672}\right)=0.011907\) dependent events

      3. \(P(\text { Green then Red })\) with replacement

        \(\frac{287}{1673} \times \frac{311}{1673}=0.031890\) independent events

      4. \(P(\text { Green then Red })\) without replacement

        \(\left(\frac{287}{1673}\right)\left(\frac{311}{1672}\right)=0.031909\) dependent events

      5. \(P(\text { Both Orange Skittles })\) with replacement

        \(\left(\frac{154}{1673}\right)\left(\frac{154}{1673}\right)=0.008473\) independent events

      6. \(P(\text { Both Orange Skittles} )\) without replacement

        \(\left(\frac{154}{1673}\right)\left(\frac{153}{1672}\right)=0.008423\) dependent events

    3. The Multiplication Rule

    • \(P(A \text { and } B)=P(A) P(B | A)\)
      •  
    • When A and B are independent events,

      \(P(A \text { and } B)=P(A) P(B)\)


  3. The table below summarizes a survey of people on their political affiliation.
    Republicans Democrats Independents Total Number
    Over 60 Years Old 46 39 1 86
    18-60 Years Old 5 9 0 14
    Total Numer 51 48 1 100
    1. Randomly select 1 person
      1. P (18-60 Years Old | Democrat)

        \(\frac{9}{48}=.1875\)

      2. P (Republican | 18-60 Years Old)

        \(\frac{5}{14}=.3571\)

    2. Randomly select 2 different people:
      1. P (Both 18-60 Years Old)

        \(\frac{14}{100} \times \frac{13}{99}=.0184\)

      2. P (Both Democrat)

        \(\frac{48}{100} \times \frac{47}{99}=.2279\)

      3. P(Democrat then Republican)

        \(\frac{48}{100} \times \frac{51}{99}=.2473\)

    3. Randomly select 3 different people
      1. P(none over 60 years old)

        \(\frac{14}{100} \times \frac{13}{99} \times \frac{12}{98}=0.0023\)

      2. P(at least one over 60 years old)

        \(1-0.0023=0.9977\)

      3. P(no Republicans)

        \(\frac{49}{100} \times \frac{48}{99} \times \frac{47}{98}=0.1139\)

    4. The 5% Guideline for Cumbersome Calculations: If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent).

  4. As of 2022, 29% of households in the United States own cats. If you pick two households at random, what is the probability they both own a cat?

    \((0.29)(0.29)=0.0841\)

    5. Probability Decisions

    The 2-way table shows the number of survey participants who received a speeding ticket last year and the color of their car.


    Red Car Not Red Car Total
    Speeding Ticket 45 36 81
    No Speeding Ticket 404 410 814
    Total 449 446 895

  5. Problem

    If one person is randomly selected from the group, what is the probability that this person drives a red car or got a speeding ticket?

    Type of Probability Problem

    One selection

    • Basic: use total sample space for total outcomes
    • Conditional: use specified row or column total for total outcomes
    • Event A or Event B: Addition Rule

    Two or more selections

    • Multiplication Rule with replacement: sample space stays the same
    • Multiplication Rule without replacement: sample space decreases after each selection

    Evidence and Result

    How did you know the type of probability problem?





    Calculations:







  6. Problem

    If the person drives a red car, what is the probability they got a speeding ticket last year?

    Type of Probability Problem

    One selection

    • Basic: use total sample space for total outcomes
    • Conditional: use specified row or column total for total outcomes
    • Event A or Event B: Addition Rule

    Two or more selections

    • Multiplication Rule with replacement: sample space stays the same
    • Multiplication Rule without replacement: sample space decreases after each selection

    Evidence and Result

    How did you know the type of probability problem?





    Calculations:





  7. Problem

    If two people are randomly selected from the group, what is the probability they both got a speeding ticket last year?

    Type of Probability Problem

    One selection

    • Basic: use total sample space for total outcomes
    • Conditional: use specified row or column total for total outcomes
    • Event A or Event B: Addition Rule

    Two or more selections

    • Multiplication Rule with replacement: sample space stays the same
    • Multiplication Rule without replacement: sample space decreases after each selection

    Evidence and Result

    How did you know the type of probability problem?





    Calculations:





  8. Problem

    What is the probability that a randomly selected driver does not have a red car?

    Type of Probability Problem

    One selection

    • Basic: use total sample space for total outcomes
    • Conditional: use specified row or column total for total outcomes
    • Event A or Event B: Addition Rule

    Two or more selections

    • Multiplication Rule with replacement: sample space stays the same
    • Multiplication Rule without replacement: sample space decreases after each selection

    Evidence and Result

    How did you know the type of probability problem?





    Calculations:





  9. Problem

    If one participant is randomly selected, what is the probability that the selected person is the owner of a red car who did not get a speeding ticket last year?

    Type of Probability Problem

    One selection

    • Basic: use total sample space for total outcomes
    • Conditional: use specified row or column total for total outcomes
    • Event A or Event B: Addition Rule

    Two or more selections

    • Multiplication Rule with replacement: sample space stays the same
    • Multiplication Rule without replacement: sample space decreases after each selection

    Evidence and Result

    How did you know the type of probability problem?





    Calculations: