At least one dry pen in the grab

Question

A drawer holds 10 pens, of which 3 are dry (out of ink) and 7 write fine. You grab 4 pens at random. What is the probability that at least one of the 4 pens is dry?

Accepted Answer

"At least one" is awkward to count directly, because it covers 1 dry, 2 dry, or 3 dry. The clean trick is to find the chance of the opposite event, "no dry pens at all," and subtract from 1. Chance of zero dry pens means all 4 come from the 7 good pens. Using combinations, where \binom{n}{k} counts ways to choose k from n: P(	ext{no dry})=\frac{\binom{7}{4}}{\binom{10}{4}}=\frac{35}{210}=\frac{1}{6}. Subtract from 1 to get "at least one dry": 1-\frac{1}{6}=\frac{5}{6}\approx 0.833. Takeaway. For "at least one," flip to the complement "none," which is a single easy count, then subtract from 1…

At least one dry pen in the grab

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