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probability

This topic has 2 expert replies and 1 member reply
soudeh Junior | Next Rank: 30 Posts
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20 Apr 2017
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probability

Post Sun Oct 15, 2017 4:46 pm
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

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Post Thu Nov 09, 2017 6:21 pm
One more approach: take it case by case!

Case 1: We pull a multiple of 6

Case 2: We DON'T pull a multiple of 6, but we pull a multiple of 2 and a multiple of 3, and 2 * 3 = 6

From here, we can split things up into subcases:

Case 1(i): We pull the multiple of 6 from the first deck
Case 1(ii): We DON'T pull the multiple of 6 from the first deck, but we do pull it from the second deck

Case 1(i) = 2/10
Case 1(ii) = (8/10) * 2/10 = 16/100

So for the first two cases, we've got 2/10 + 16/100, or 36/100, or .36.

Case 2(i): We pull the multiple of 2 from the first deck and the multiple of 3 from the second deck
Case 2(ii): We pull the multiple of 3 from the first deck and the multiple of 2 from the second deck

Remember here too that the multiples of 2 and of 3 that we pull CANNOT be multiples of 6. (We already treated all those scenarios in Case 1.) So our multiples of 2 are 14, 16, and 20, and our only multiple of 3 is 15. That gives us

Case 2(i): 3/10 * 1/10 = 3/100
Case 2(ii): 1/10 * 3/10 = 3/100

That adds another 6/100, or .06, to our probability.

From here, we've got .36 + .06, or .42.

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soudeh Junior | Next Rank: 30 Posts
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Post Mon Oct 16, 2017 2:44 pm
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.
Amazing explanation.... thanks a lot Smile

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Post Mon Oct 16, 2017 12:28 pm
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.

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Ceilidh Erickson
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EdM in Mind, Brain, and Education
Harvard Graduate School of Education


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