Unit 2 Probability

4.1 Probability Distributions

Human blood comes in different types. Each person has a specific ABO type (A, B, AB, or O) and Rh factor (positive or negative). Hence, if you are O+, your ABO type is O and your Rh factor is positive. A probability model for blood types in the United States is given:

  1. Blood Type A+ A- B+ B- AB+ AB- O+ O-
    Probability 0.357 0.063 0.085 0.015 0.034 0.006 0.374 0.066
    1. What is the probability that a randomly chosen person has blood type A?

      2. \(P(A+\text { or } A-)=P(A+)+P(A-)=0.357+0.063=0.420\)

      Hence, roughly 42% of residents in the U.S. have ABO-type A blood.

    2. What is the probability that two randomly chosen U.S. residents both have type A blood?

      3. \(\text { P(Person 1 is type A and Person 2 is type A) = }(0.420)(0.420)=0.1764\)

      Hence, there is roughly an 18% chance that two randomly selected U.S. residents will both have type A blood.

    3. What is the probability that at least one of the two people does not have type A blood?

      4. \( \text {P(at least one not A)}=1-P(\text { both A } )=1-0.1764=0.8236\)

      Therefore, in random samples of size two, we expect at least one of the two people not to be type A about 82% of the time.

    A probability distribution is what will probably happen, not necessarily what actually happened. Probability distributions are theoretical, not experimental. These distributions are usually given in the form of a graph, table or formula.

    A random variable, x, represents a value associated with each outcome of the distribution.

    Qualifiers for probability distributions:
    • \(\sum P(x)=1\)
    • \(0 \leq P(x) \leq 1\)     All probabilities are between impossible and certain, inclusive.
  2. Do the following distributions meet both requirements? Why/Why not?

    Distribution A:

    x values Probability P(x)
    2 0.4
    3 1.2
    5 0.1

    Distribution B:

    x values Probability P(x)
    2 0.4
    3 0.4
    5 0.2

    Distribution C:

    x values Probability P(x)
    2 0.1
    3 0.6
    5 0.1

    Distribution A is not a probability distribution. It does not meet either qualification.

    Distribution B is a probability distribution. It meets both qualifications.

    Distribution C is not a probability distribution. It does not meet the first qualification.

    3. Two Types of Random Distributions: Discrete and Continuous
    Discrete Distributions:

    A random variable is discrete when it has a finite or countable number of possible outcomes that can be listed. A discrete distribution lists each possible outcome for x together with the probability of x occurring. The probability distribution is a relative frequency distribution.

    Continuous Distributions:

    A random variable is continuous when it has an uncountable number of possible outcomes, represented by an interval on a number line.

  3. Determine whether each random variable, x, is discrete or continuous. Explain your reasoning.
    1. Let x represent the number of cars in the PSCC Hardin Valley parking lots.
      2. Discrete
    2. Let x represent the height of a random student on PSCC campus.
      3. Continuous
    Creating a Discrete Probability Distribution
  4. You take a 2 question quiz. The first question, Q1, is T/F and the second question, Q2, is multiple choice with a, b, c, or d as your choices. You guess on both. After you submit your quiz, what are the following probabilities?
    1. P (Q1 is correct) = 1/2 = 0.5
    2. P (Q2 is correct) = 1/4 = 0.25
    3. P (Both are Correct) = P (C and C ) = (1/2)(1/4) = 1/8 = .125
    4. P (C and W) = (1/2)(3/4) = 3/8 = .375
    5. P (W and C) = (1/2)(1/4) = 1/8 = .125
    6. P (W and W) = (1/2)(3/4) = 3/8 = .375
    7. Use the information in a-f to create a probability distribution table:

      Let x = number correct

      x values Probability P(x)
      0

      P(W and W) = 0.375
      1

      P(C and W) + P(W and C) = 0.5
      2

      P(C and C) = 0.125
    8. Does our distribution meet both requirements for a probability distribution?

      \(\sum P(x)=1\) Yes, the probabilities add to 1.

      \(0 \leq P(x) \leq 1\) Yes, all the probabilities are less than 1 and positive.

  5. Use the data in the table below to answer the questions
    y 1 2 3 4 5 6 7 8
    p(y) 0.15 0.23 0.19 0.23 0.12 0.05 0.02 0.01
    1. What is the probability that a randomly selected family will consist of at least two people?

      2. \(P(y \geq 2)=1-P(y\lt2)=1-p(1)=0.85\)

    2. What is the probability that a randomly selected family will have two to four members?

      3. \(\mathrm{P}(2 \leq \mathrm{y} \leq 4)=\mathrm{P}(\mathrm{y}=2 \text { or } \mathrm{y}=3 \text { or } \mathrm{y}=4)=\mathrm{p}(2)+\mathrm{p}(3)+\mathrm{p}(4)=0.65\)

    3. What is the mean family size?

      4. \(\mu=3.22\)

    4. What is the standard deviation for family size?

      5. \(\sigma \approx \sqrt{2.512} \approx 1.58\)

    5. Would selecting a family with 8 members be unusual?

      6. \(\text { The minimum usual value is } \mu-2 \sigma=3.22-2(1.58) \text { which is } 0.06\)

      \(\text { The maximum usual value is } \mu+2 \sigma=3.22+2(1.58) \text { which is } 6.38.\)

      So, yes families with 7 or 8 members are unusual.

  6. A cell phone service has the following distribution of cell phones associated with each friends and family account.

    (The symbol 0+ denotes a positive probability value that is very small. We are not willing to say this is an impossible event. Use P(x=6) = 0 in your calculations.)

    x values Probability P(x)
    1 .02
    2 .13
    3 .32
    4 .39
    5 .14
    6 0+

    If one account is randomly selected, find the following probabilities for the number of phones associated with the account:

    1. Probability of exactly 3 phones:

      P(x = 3) = .32

    2. Probability of 4 or more phones:

      P(x ≥ 4) = P(x = 4) + P(x = 5) + P(x = 6) = .39 + .14 + 0 = .53

    3. Probability of at least 2 phones:

      P(x ≥ 2) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = .13 + .32 + .39 + .14 + 0 = .98

      Or use the complement rule: 1 – P(x = 1) = 1 - .02 = .98

    4. Probability of at most 3 phones:

      P(x ≤ 3) = .02 + .13 + .32 = .47

    5. Probability of no more than 2 phones:

      P(x ≤ 2) = .02 + .13 = .15

    6. Probability of between 2 and 5 phones:

      P(2 < x < 5) = P(x = 3) + P(x = 4) = .32 + .39 = .71

    7. Probability of between 2 and 5 phones, inclusive:

      P(2 ≤ x ≤ 5) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) = .13 + .32 + .39 + .14 = .98

    7. Mean (Expected Value) and Standard Deviation of a Discrete Probability Distribution

  7. A cell phone service has the following distribution of cell phones associated with each friends and family account.
    x values Probability P(x)
    1 .02
    2 .13
    3 .32
    4 .39
    5 .14
    6 0+
    1. Did we keep both rules for probability distributions? Yes

      \(\sum P(x)=1\)

      \(0 \leq P(x) \leq 1\)

    2. MEAN (EXPECTED VALUE): \(\mu=E(x)=\sum[x * P(x)]\)

      \(\mu=[1 * .02]+[2 * .13]+[3 * .32]+[4 * .39]+[5 * .14]+[6 * 0]=3.5\)

    3. VARIANCE:

      \( \sigma^2=(.95)^2=.9025=.90\)

    4. STANDARD DEVIATION:

      \(\sigma=\sqrt{\sigma^{2}}=\sqrt{.91}=.95\)

    5. UNUSUAL VALUES:

      Min usual value: \(\mu-2 \sigma = \) \(3.5-2(.95)=1.6\)

      Max usual value: \(\mu+2 \sigma = \) \(3.5+2(.95)=5.4\)

      Would selecting an account with 1 phone be unusual? Yes, 1 is less than the min usual value.

      Would selecting an account with 2 phones be unusual? No, 2 is within the usual range.

      Would selecting an account with 5 phones be unusual? No, 5 is within the usual range.