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What will be the probability that all three defective pills will be selected in the case below?



A batch of pills consists of 7 that are good and 3 that are defective.  Three pills are randomly selected and tested.a) How many different samples of pills are possible when three pills are randomly selected (without replacement) from the 10 that are available?.


Asked by Manello, Last updated: Apr 15, 2024

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b.Lisa

b.Lisa

b.Lisa
B.Lisa

Answered Feb 26, 2018

There are seven distinct ways to draw pills without replacing: good, bad, good; good, good, good; bad, bad, bad; good, good, bad; good, bad, bad; and so on. That said, when it comes to individual pills, there are 720 different distinct ways to draw the pills without replacing the ones already drawn. The odds of drawing all three defective pills in a single round are 0.001388889. To get the different ways of drawing without replacing, you multiply 10 by 9 by 8. For the odds, you multiply 1/10 by 1/9 by ⅛, subtracting one from the number each time.


These odds are not in anyone’s favor, since at least two of the defective pills can slip through and not get tested randomly. In fact, it’s more probable that the three pills drawn and tested will be the good pills, and no defective pills are drawn.

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