There are two decks of 10 cards each. The cards in each deck are labeled with integers from 11 to 20 inclusive. If we pick a card from each deck at random, what is the probability that the product of the numbers on the picked cards is a multiple of 6? A. 0.23 B. 0.36 C. 0.40 D. 0.42 E. 0.46

(D) Let’s consider an outcome to be an ordered pair (x, y), where x is selected from the deck #1 and y is selected from the deck #2. There are 10 × 10 = 100 such possible outcomes (pairs). Let’s count the number of favorable outcomes.

First, factorize the given numbers. 11 is prime 12 = 2 × 2 × 3 13 is prime 14 = 2 × 7 15 = 3 × 5 16 = 2 × 2 × 2 × 2 17 is prime 18 = 2 × 3 × 3 19 is prime 20 = 2 × 2 × 5

The product will be a multiple of 6 (6 = 2 × 3) in each of the following cases: - one of the picked numbers is divisible by 6, which means it has 2 and 3 in its factorization The numbers in the decks that are divisible by 6 are 12 and 18. There are pairs (12, y), (18, y), (x, 12), (x, 18). Each pair alone gives us 10 favourable outcomes, but outcomes (12, 18), (12, 12), (18,18), (18, 12) are counted twice. So this case gives us 4 × 10 – 4 = 36 favourable outcomes.

- one of the picked numbers is divisible by 2, while the other one is divisible by 3 In this case we do NOT consider numbers that are divisible by 6, because we have already counted those. The numbers that are divisible by 2 are 14, 16, 20. There is only one number which is divisible by 3. It is 15. Therefore this case gives us (3 × 1) × 2 = 6 favourable outcomes.

There are 36 + 6 = 42 favourable outcomes among 100 possible outcomes. Therefore the probability is 42/100 = 0.42. The correct answer is D. ---------- I think the question should say that the decks are non-identical specifically.

I think the question should say that the decks are non-identical specifically.

No, there is no necessity in this. If you have doubts regarding why we use ordered pairs, then imagine that one of the decks lies to the left and one to the right.

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