probability

ceilidh.erickson wrote:Problems like this are labor-intensive, because we have to count a number of different possibilities. To get a product that is a multiple of 6, we must have a factor of 2 and a factor of 3 between the two numbers. We could have a multiple of 6 from the 1st deck with a prime from the second, a multiple of 2 from the first and a multiple of 3 from the second... etc.

There might be more elegant ways to make sure we've counted all possibilities, but with a finite list like this, I prefer to just list out all of the possibilities. I'll start by listing every card from deck 1, and then every card that could then be pulled from deck 2 to get a multiple of 6. If the card from deck 1 is already a multiple of 6, then any card from deck 2 will work. If the Deck 1 card is a multiple of 3, we must pull a multiple of 2 from Deck 2 (and vice versa). If the Deck 1 card is not a multiple of 2 or 3, we must pick a multiple of 6 from Deck 2.

Deck 1_____Deck 2____# of multiples of 6:
11 ----> 12, 18 -----------------> 2
12 ----> all -------------------> 10
13 ----> 12, 18 -----------------> 2
14 ----> 12, 15, 18 -------------> 3
15 ----> 12, 14, 16, 18, 20 ----> 5
16 ----> 12, 15, 18 -------------> 3
17 ----> 12, 18 -----------------> 2
18 ----> all -------------------> 10
19 ----> 12, 18 -----------------> 2
20 ----> 12, 15, 18 -------------> 3

Now we just add these up:
Total # of combinations that would yield a multiple of 6 = 42

Total possible # of combinations with 10 cards in each deck = 10*10 = 100

Therefore, the probability is 42/100 ---> D.