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Math Help - The Binomial Distribution

  1. #1
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    The Binomial Distribution

    can anyone help?


    A circuit has 6 breakers. The probability that each breaker will fail is
    0.1. If the circuit is activated, .find the probability that exactly two
    breakers will fail. Each breaker is independent of any other breaker.
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  2. #2
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    Quote Originally Posted by jickjoker View Post
    can anyone help?
    A circuit has 6 breakers. The probability that each breaker will fail is
    0.1. If the circuit is activated, .find the probability that exactly two
    breakers will fail. Each breaker is independent of any other breaker.
    You should understand that this is not a tutorial service.
    You can look it up.
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  3. #3
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    Quote Originally Posted by jickjoker View Post
    can anyone help?



    A circuit has 6 breakers. The probability that each breaker will fail is
    0.1. If the circuit is activated, .find the probability that exactly two
    breakers will fail. Each breaker is independent of any other breaker.
    What have you tried? Where are you stuck?
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  4. #4
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    just trying to verify that i have worked it out correctly, i like to do brainteazers in my spare time and this
    one come up , i think the answer is

    6c2 , 0.1 *2 , 0.9 *4 = 0.098415
    9.8% chance
    i think thats correct
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  5. #5
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    Quote Originally Posted by jickjoker View Post
    just trying to verify that i have worked it out correctly, i like to do brainteazers in my spare time and this
    one come up , i think the answer is
    6c2 , 0.1 *2 , 0.9 *4 = 0.098415
    9.8% chance
    i think thats correct
    That is correct: \binom{6}{2}(.1)^2(.9)^4=0.098415
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