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Math Help - Probability Counting Rules

  1. #1
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    Probability Counting Rules

    A business office orders paper supplies from one of three vendors, v1, v2, or v3. Orders are randomly to be placed to those three vendors on two successive days, one order per day. Thus (v2, v3) might denote that vendor v2 gets the order on the first day and vendor v3 gets the order on the second day.

    a) Describe the sample space of this experiment in ordered pairs on two successive days.
    b) Assume the vendors are selected at random to fill the order. Assign a probability to each elementary outcome.
    c) Let A denote the event that the same vendor gets both orders and B the event that the second vendor, v2, gets at least one order. Find P(A), P(B), P(AB), and P(AUB).

    My answers:
    a) S = {(v1, v1), (v1, v2), (v1, v3), (v2, v1), (v2, v2), (v2, v3), (v3, v1), (v3, v2), (v3, v3)}
    b) The probability of each outcome is 1/9.
    c) P(A) = 1/3. P(B) = 5/9. P(AB) = (1/3)(5/9) = 5/27. P(AUB) = 1/3 + 5/9 - 1/9 = 7/9.

    Will you please give me feedback for these answers?
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  2. #2
    Grand Panjandrum
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    Re: Probability Counting Rules

    Quote Originally Posted by lovesmath View Post
    A business office orders paper supplies from one of three vendors, v1, v2, or v3. Orders are randomly to be placed to those three vendors on two successive days, one order per day. Thus (v2, v3) might denote that vendor v2 gets the order on the first day and vendor v3 gets the order on the second day.

    a) Describe the sample space of this experiment in ordered pairs on two successive days.
    b) Assume the vendors are selected at random to fill the order. Assign a probability to each elementary outcome.
    c) Let A denote the event that the same vendor gets both orders and B the event that the second vendor, v2, gets at least one order. Find P(A), P(B), P(AB), and P(AUB).

    My answers:
    a) S = {(v1, v1), (v1, v2), (v1, v3), (v2, v1), (v2, v2), (v2, v3), (v3, v1), (v3, v2), (v3, v3)}
    b) The probability of each outcome is 1/9.
    c) P(A) = 1/3. P(B) = 5/9.
    You are Ok up to here


    P(AB) = (1/3)(5/9) = 5/27.

    P(AB) is the probability that the same vendor gets both orders and that v2 gets at least one. There is only one point for which this is true (v2,v2)

    P(AUB) = 1/3 + 5/9 - 1/9 = 7/9.
    OK

    but since

    P(AUB)=P(A)+P(B)-P(AB)

    why do you have 1/9 for P(AB) here when you have something wrong when asked for it directly?

    CB
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  3. #3
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    Re: Probability Counting Rules

    I'm not sure why I multiplied to get P(AB); it is obviously 1/9. After plugging in P(A), P(B) and P(AB), P(AUB) should still equal 7/9.
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  4. #4
    Grand Panjandrum
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    Re: Probability Counting Rules

    Quote Originally Posted by lovesmath View Post
    I'm not sure why I multiplied to get P(AB); it is obviously 1/9. After plugging in P(A), P(B) and P(AB), P(AUB) should still equal 7/9.
    Yes

    CB
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