Question 477798
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If you have 14 bad calculators and 24 good calculators and you pull 4, 
what is the probability that 1 is bad ?
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        The solution in the post by @Theo is incorrect.

        It is because @Theo used the Binomial distribution model in his solution, 

        but this model does not work in this case.


        See my correct solution below.



<pre>
We should calculate the total number of all possible different quadruples,
the number of all possible different quadruples containing 1 bad and 3 good calculators
and relate the second quantity to the first quantity.


The total number of all calculators in the problem is 14+24 = 38.



The total number of all possible quadruples of 38 calculators is the number

of combinations  C(38,4) = {{{(38*37*36*35)/(1*2*3*4)}}} = 73815.



The total number of all possible quadruples containing 1 bad and 3 good calculators 
is this product

    C(14,1)*C(24,3) = 14*2024 = 28336.



Finally, the probability under the problem's question is

    P = {{{28336/73815}}} = 0.3839  (rounded).    <U>ANSWER</U>
</pre>

Solved correctly.