Question
A shipment of 20 keyboards contains 3 defective units. A company buys four of these units. What is the probability of the company buying at least three nondefective units?
Step 1
This can be done using the combination formula which is given by $\frac{n!}{r!(n-r)!}$ where n is the total number of items, and r is the number of items to choose. So, the total number of ways is $\frac{20!}{4!(20-4)!} = 4845$. Show more…
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