Find the probability that one ink pen was dried out a store clerk collected seven dried-out ink pens from the store's cash registers. while talking with a customer, she accidentally dropped the pens into a basket of thirty ink pens which were for sale. another customer walked by and picked up three ink pens to purchase. Find the probability that one ink pen was dried out. Okay, so i put (C(3,1)C(34,6)) /C(37,7) = .3919 which is ironically the correct answer. however, I still got it wrong because it was the wrong work? i forgot what the correct work was ( i know... i should have wrote it down); but i was hoping someone could explain to me why it is wrong?

Since the customer chooses 3 pens to purchase, your denominator should be $C(37,3)$. That's the total number of ways they could make their choice. To count the ways they could pick one dry one and two good ones, we just write $C(7,1)C(30,2)$ - choose one of the seven dry ones, and two of the thirty good ones. That's the numerator.

Does that way of thinking about it help at all? We're counting, out of all the ways to purchase $3$ pens, how to pick $1$ from the dry set, and $2$ from the good set.

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Your way also works, because you're dividing the pens into the ones purchased and the ones not purchased - $3$ and $34$ - and counting the ways that the seven dry pens could show up as one in the purchased set, and six in the non-purchased set. That's a sideways, but equivalent approach. In symbols:

$$\Large{\frac{\binom{k}{x}\binom{N-k}{n-x}}{\binom{N}{n}}=\frac{\binom{n}{x}\binom{N-n}{k-x}}{\binom{N}{k}}}$$