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Math Help - Probably 2 very easy probability problems

  1. #1
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    Probably 2 very easy probability problems

    Hi, I need help understanding 2 probability problems.

    I have a box with 24 circuit boards and I know 5 are faulty, if I take out 6 boards what is the probability that one of those 6 is faulty?

    I am unsure about the second one althought it seems very easy, almost to easy.

    A medicine makes 0.07 sick, what is the probability that 3 out of 10 people that take it will get sick?

    I have a hard time following the examples in the book so I hope you can shed some light on what seems to be 2 easy questions.

    Thanks!
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  2. #2
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    The second one is a binomial probability.

    C(10,3)(\frac{7}{100})^{3}(\frac{93}{100})^{7}
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  3. #3
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    The first one is a hypergeometric probability,

    You choose 1 from the 5 faulty ones and 5 from the 19 good ones.

    \frac{C(5,1)C(19,5)}{C(24,6)}
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  4. #4
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    Thank you but can you explain how to proceed from there?
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  5. #5
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    Run it through a calculator. C(10,3), for instance, means the number ways to choose 3 items out of 10. C(10,3)=120

    The formula is \frac{n!}{(n-k)!k!}=\frac{10!}{(10-3)!3!}=120
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  6. #6
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    Ok now I feel even dumber, when I run that I get 4320.

    Also, how do i get the % of 3 out of 10 if I even get 120?

    Sorry to be such a bother.
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  7. #7
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    Can I get these answers checked please?

    Question 1 = 5*19*18*17*16*15*6*5*4*3*2*1 / (1*5*4*3*2*1*24*23*22*21*20*19*18) = 0.02400 = 2.4%


    Question 2 = 10*9*8/3/2/1*0.07^3 * (1-0.07)^7 = 0.02477 = 2.5%

    Thanks.
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