Question 1183656
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A batch of pills consist of 9 good pills and 3 that are defective (contain the wrong amount of the drug). 
If 4 pills are randomly selected without replacement, what is the probability 
that all 3 of the defective pills are selected?
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We should calculate the number of the quadruples that contain 3 defective pills and 1 good pill.


Then we should relate this number to the number of all possible quadruples of 9+3 = 12 pills.


The numerator of the fraction is  {{{C[3]^3*C[9]^1}}} = 1*9 = 9.


The denominator of the fraction is  {{{C[12]^4}}} = {{{(12*11*10*9)/(1*2*3*4)}}} = 495.


The final probability under the problem's question is  {{{9/495}}} = {{{1/55}}} = 0.01818 = 1.818%  (rounded).    <U>ANSWER</U>
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Solved.