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Math Help - probability distribution help

  1. #1
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    probability distribution help

    In a lot of 12 washing machines , there are 3 defective pieces . A person has ordered 4 washing machines . Find the probability that all the four are good .

    My working :

    X-B(4 , 0.75)
    P(X=4) = 4C4(0.75)^4 (0.25)^0 = 81/256

    But my answer is wrong . Where is my mistake again ? Thanks
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    In a lot of 12 washing machines , there are 3 defective pieces . A person has ordered 4 washing machines . Find the probability that all the four are good .

    My working :

    X-B(4 , 0.75)
    P(X=4) = 4C4(0.75)^4 (0.25)^0 = 81/256

    But my answer is wrong . Where is my mistake again ? Thanks
    X is not binomial because you're choosing without replacement (in fact, X is hypergeometric).

    A combinatorial approach (which is what the hrpergemetric distribution is actually based on):

    \Pr(X = 4) = \frac{^9C_4 \cdot ^3C_0}{^{12}C_4}.

    Do you see where each bit comes from?
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    In a lot of 12 washing machines , there are 3 defective pieces . A person has ordered 4 washing machines . Find the probability that all the four are good .

    My working :

    X-B(4 , 0.75)
    P(X=4) = 4C4(0.75)^4 (0.25)^0 = 81/256

    But my answer is wrong . Where is my mistake again ? Thanks
    As Mr. Fantastic (and he really is!) said, this is sampling without replacement and the formula you used applies only to sampling with replacement.

    It's better to think like this: there are, to start with, 12 machines, 9 of which are good. The probability that the first machine selected is good is 9/12= 3/4. Assuming that happens, there are now 11 machines, 8 of which are good. The probability that the second machine selected is good is 8/11. Assuming that happens, there are now 10 machines, 7 of which are good. The probability the third machine is selected is 7/10. Finally, we have 9 machines left, 6 of which are good and the probability that the fourth machine selected is good is 6/9= 2/3. We can assume each selected was good because we are only calculating the probability that all four selected are good: it is (9/12)(8/11)(7/10)(6/9)= (3/4)(8/11)(7/10)(2/3)
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  4. #4
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    Re:

    Oh oh , i know my mistake . Thanks a lot , Mr F and Hallsofivy for that effort .
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