A batch of pills consists of 7 that are good and 3 that are defective (because they contain the wrong amount of the drug). a. How many different permutations are possible when all 10 pills are randomly selected (without replacement)? b. If 3 pills are randomly selected without replacement, find the probability that all three of the defective pills are selected.
Added by Marcos H.
Step 1
Given that there are 10 pills in total, with 3 defective and 7 good pills, we have: \(n = 10\) and \(r = 10\). Therefore, the number of permutations possible is: \(10P10 = \frac{10!}{(10-10)!} = \frac{10!}{0!} = 10! = 3628800\). Show more…
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