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Math Help - Waiting line probabilities

  1. #1
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    Waiting line probabilities

    Probabilities:
    Number of Customers in the System ---- Probability
    ----------------------------------------------------------------
    0 ---- 0.5600
    1 ---- 0.2464
    2 ---- 0.1084
    3 ---- 0.0477
    4 ---- 0.0210
    5 ---- 0.0092
    6 OR MORE ---- 0.0073

    The probability that a customer will have to wait is 0.4400. Thus the probability that that will not is 0.5600.

    (1) What is the probability that 1 customer is receiving a haircut and no one is waiting?
    I figured it like this: P(not waiting) = 0.5600
    P(1 customer in the system) = 0.2464
    P(not waiting ∩ 1 customer in the system) = 0.5600 * 0.2464 = 0.137984 <--- Does that seem right?

    (2) The probability of 2 customers being in the system is .1084. The probability that they will have to wait is .4400, and therefore the probability that they will NOT have to wait is .5600. What is the probability that 2 customers are in the system, and 1 is waiting and 1 is NOT waiting?

    (3) The probability of 3 customers being in the system is .0477. The probability that they will have to wait is .4400, and therefore the probability that they will NOT have to wait is .5600. What is the probability that 3 customers are in the system, and 2 are waiting and 1 is NOT waiting?

    THANK YOU!
    Last edited by chloe625; November 29th 2008 at 12:20 PM.
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  2. #2
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    You might want to include more information about the question. Assuming there is only one server, the answer to (1), (2) and (3) is just the probability that there are 1, 2 and 3 customers in the system respectively.
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