SOLUTION: A batch of pills consist of 10 good pills and 4 that are defective (contain the wrong amount of the drug). If 5 pills are randomly selected without replacement, what is the probabi

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Question 1183758: A batch of pills consist of 10 good pills and 4 that are defective (contain the wrong amount of the drug). If 5 pills are randomly selected without replacement, what is the probability that all 4 of the defective pills are selected?
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
There are 14 pills and 14C5 ways to choose or 2002 ways.
5 ways to choose 4 pills and the last pill is always good, so 5/2002=0.00250 probability.
5C4*1/14C5
also, (10/14) ways for the first (4/13)(3/12)(2/11)(1/10) for the rest, and 5 ways that can happen with same numerator/denominator, each =240/240240, and that is the same probability.
Answer by ikleyn(52832)   (Show Source): You can put this solution on YOUR website!
.
A batch of pills consist of 10 good pills and 4 that are defective (contain the wrong amount of the drug).
If 5 pills are randomly selected without replacement, what is the probability that all 4 of the defective pills are selected?
~~~~~~~~~~~~~~~~


            I do not agree with the solution by @Boreal.
            My solution and my answer are different. 


The number of all possible subsets of 5 pills, randomply selected from the total set of 10+4 = 14 pills is

     =  = 2002.


It is the cardinality of the total space of events.


The number of all possible subsets of 5 pills, containing 4 defective pills (and, hence, one good pill), is 10.

These subsets contain all 4 defective pills and one of 10 good pills, so the cardinality of this set 
("favorable" set of events) is 10.


The probability under the problem's question is thus


    P =  =  =  = 0.004995  (rounded).    ANSWER


If you want a general formula, here it is  P = .

Solved and explained.



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