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Math Help - A probability with two conditions

  1. #1
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    A probability with two conditions

    Can someone please help me with this question? I *think* my logic is correct but I always get the wrong final answer. =/

    Question: Given the following information concerning how many days a person has to wait for a repair:

    (# of days, probability)
    (1, 0.10) *means 10% chance to wait 1 day*
    (2, 0.25)
    (3, 0.30)
    (4, 0.25)
    (5, 0.10)

    A random sample of 5 customers selected, what is the probaility that at least 4 people will have to wait more than 3 days?

    I was trying to solve it by adding the probability of 4 people waiting more than 3 days to 5 people waiting more than 3 days, in other words P(4 people, day = 4, 5) + P(5 people, day = 4, 5) = [(5 select 4)(0.25 + 0.10)^4(1 - 0.25 - 0.10)^1] + [(5 select 5)(0.25 + 0.10)^5(1 - 0.25 - 0.10)^0], but that answer is wrong.
    (sorry I'm not sure how to post those math symbols in picture format)
    Can someone please enlighten me on this, thanks!
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  2. #2
    MHF Contributor
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    Re: A probability with two conditions

    Quote Originally Posted by Selena View Post
    Can someone please help me with this question? I *think* my logic is correct but I always get the wrong final answer. =/

    Question: Given the following information concerning how many days a person has to wait for a repair:

    (# of days, probability)
    (1, 0.10) *means 10% chance to wait 1 day*
    (2, 0.25)
    (3, 0.30)
    (4, 0.25)
    (5, 0.10)

    A random sample of 5 customers selected, what is the probability that at least 4 people will have to wait more than 3 days?

    I was trying to solve it by adding the probability of 4 people waiting more than 3 days to 5 people waiting more than 3 days, in other words P(4 people, day = 4, 5) + P(5 people, day = 4, 5) = [(5 select 4)(0.25 + 0.10)^4(1 - 0.25 - 0.10)^1] + [(5 select 5)(0.25 + 0.10)^5(1 - 0.25 - 0.10)^0], but that answer is wrong.
    (sorry I'm not sure how to post those math symbols in picture format)
    Can someone please enlighten me on this, thanks!
    You could alternatively calculate the probability of

    1 person waiting less than 4 days
    or nobody waiting less than 4 days.

    See if the answers match up.
    Last edited by Archie Meade; July 19th 2011 at 01:56 PM.
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  3. #3
    Junior Member
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    Re: A probability with two conditions

    Hmm why do I need to calculate 1 person waiting less than 4 days or nobody waiting less than 4 days? Those aren't complements to the two conditions given in the question right?
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  4. #4
    Junior Member
    Joined
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    Re: A probability with two conditions

    LOL nevermind my solution is actually right. Minor calculation error with the selection!
    Thanks anyways lol.
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