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Math Help - simple i think

  1. #1
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    simple i think

    We have 60 TV's on the shop , 5 of them have defects.
    if we buy 2 TV 's how is the probability that both have defects ?
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  2. #2
    Kai
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    Yes, simply, prob to get 1 defective tv =5/60

    Ermm my mistake, binomial distribution applies here....

    http://personal.maths.surrey.ac.uk/s...poiss_bin.html

    Some hints there
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by beq!x View Post
    We have 60 TV's on the shop , 5 of them have defects.
    if we buy 2 TV 's how is the probability that both have defects ?
    You assume that the 60 TV's in the shop are typical so that the proportion of defectives is 1/12, and the two TV's bought are unrelated in the defectivness department.

    Then the probability that the two TV are both defective is then (1/12)(1/12).

    CB
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  4. #4
    Kai
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    Yes, gave the same answer as CB, then totally gone mad and deleted it...
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  5. #5
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    what is CB ?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by beq!x View Post
    what is CB ?
    It is CaptainBlack's initials.

    CB
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    You assume that the 60 TV's in the shop are typical so that the proportion of defectives is 1/12, and the two TV's bought are unrelated in the defectivness department.

    Then the probability that the two TV are both defective is then (1/12)(1/12).

    CB
    But the chance of you buying one defective TV is 5/60.
    Then if you buy another tv, the chances of it being defective are 4/59.
    5/60*4/59= 20/3540 = 1/177...



    You're assuming independence, are you not?
    Which they're not...
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  8. #8
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    but what if i buy 2 TV's in the same time ?
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  9. #9
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    Quote Originally Posted by Unenlightened View Post
    But the chance of you buying one defective TV is 5/60.
    Then if you buy another tv, the chances of it being defective are 4/59.
    5/60*4/59= 20/3540 = 1/177...



    You're assuming independence, are you not?
    Which they're not...
    That's not the way I read the question. The way I read it is we have 60 TV's in our shop 5 defective, we buy two more, what is the probability that these two new stock TV's are defective.

    Now if the two "we" 's in the question refer to different actors then yes the answer is different.

    CB
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  10. #10
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    Quote Originally Posted by Unenlightened View Post
    But the chance of you buying one defective TV is 5/60.
    Then if you buy another tv, the chances of it being defective are 4/59.
    5/60*4/59= 20/3540 = 1/177...



    You're assuming independence, are you not?
    Which they're not...
    so this is the right answer ?
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    That's not the way I read the question. The way I read it is we have 60 TV's in our shop 5 defective, we buy two more, what is the probability that these two new stock TV's are defective.

    Now if the two "we" 's in the question refer to different actors then yes the answer is different.

    CB
    Ah..... that's not how I read it...
    I see..

    Quote Originally Posted by beq!x View Post
    so this is the right answer ?
    That would depend, on what the question is...
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  12. #12
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    Quote Originally Posted by beq!x View Post
    We have 60 TV's on the shop , 5 of them have defects.
    if we buy 2 TV 's how is the probability that both have defects ?
    My interpretation is that you're buying two TV's from the pool of 60 TV's.

    So the probability of both being defective will be \frac{^5C_2 \cdot ^{55}C_0}{^{60}C_2}. You can think about where the different bits of this answer come from, and I'll let you simplify it.

    Alternatively, you can use the hypergeometric distribution if you know it.

    Quote Originally Posted by beq!x View Post
    what is CB ?
    Quote Originally Posted by CaptainBlack View Post
    It is CaptainBlack's initials.

    CB
    And I thought it stood for Citizen Band. Or was that Citizen Banned .... ?
    Last edited by mr fantastic; April 25th 2009 at 06:53 PM. Reason: Merged posts
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