Probability Problem !!!

[Shalini Suresh]

There are 2 decks of 10 cards each numbered from 11 to 20 inclusive. If we pick a card from each pile, what is the probablity that the product of the numbers picked is divisible by 6?

A)0.23
B)0.36
c)0.40
D)0.42
E)0.46

Answer is D

Request someone to explain .

[GMATGuruNY]

There are 2 decks of 10 cards each numbered from 11 to 20 inclusive. If we pick a card from each pile, what is the probablity that the product of the numbers picked is divisible by 6?

A)0.23
B)0.36
c)0.40
D)0.42
E)0.46

Answer is D

Request someone to explain .

Total number of products:
Number of options from deck 1 = 10.
Number of options from deck 2 = 10.
To combine these options, we multiply:
10*10 = 100.

Multiples of 6:
For the product to be a multiple of 6, it must be a divisible by 2 and 3.

Case 1: An odd value from deck 1 that is not a multiple of 3 must be multiplied by a multiple of 6.
In deck 1, odd values that are not multiples of 3 are 11, 13, 17, and 19, yielding 4 options.
In deck 2, multiples of 6 are 12 and 18, yielding 2 options.
To combine these options, we multiply:
4*2 = 8.

Case 2: An even value from deck 1 that is not a multiple of 6 must be multiplied by a multiple of 3.
In deck 1, even values that are not multiples of 6 are 14, 16, and 20, yielding 3 options.
In deck 2, multiples of 3 are 12, 15 and 18, yielding 3 options.
To combine these options, we multiply:
3*3 = 9.

Case 3: An odd multiple of 3 from deck 1 must be multiplied by a multiple of 2.
In deck 1, the only odd multiple of 3 is 15, yielding 1 option.
In deck 2, even values are 12, 14, 16, 18 and 20, yielding 5 options.
To combine these options, we multiply:
1*5 = 5.

Case 4: A multiple of 6 from deck 1 can be multiplied by any value deck 2.
In deck 1, multiples of 6 are 12 and 18, yielding 2 options.
In deck 2, there are 10 values.
To combine these options, we multiply:
2*10 = 20.

(multiples of 6)/(total number of products) = (8+9+5+20)/(100) = 42/100 = .42.

The correct answer is D.

[Ian Stewart]

There are a few ways to break down the cases here, some a bit faster than what I'll do below, but you can essentially just enumerate all of the possibilities to get an answer. We have 10 choices for the first card, and 10 for the second, so we have 100 choices in total - that's our denominator. Now, to get a product which is a multiple of 6, one of our numbers needs to be a multiple of 2, and one (possibly the same number) needs to be a multiple of 3. So, considering what we might pick with our first card:

* if we pick an odd prime (11, 13, 17, 19), we'll need to pick a multiple of 6 with the second card, so we need to pick 12 or 18, for a total of 8 possibilities (just multiplying the number of choices we have for each card)
* if we pick 12 or 18, we can pick anything with the second card, for a total of 20 possibilities
* if we pick an even number not divisible by 3 (i.e. 14, 16, or 20), we need to pick a multiple of 3 with the second card (12, 15 or 18) for a total of 9 possibilities
* if we pick 15, we need to pick one of the five even numbers with the second card, for a total of 5 possibilities

and adding the results from each case, there are 42 ways to pick numbers that give a product divisible by 6, so the answer is 42/100.

[Shalini Suresh]

Thank u Ian and Mitch !!

[subhash.sonu39]

Please post some new probability problems, I want to solve them but each I found seems so discussed and solved, if you have any more to discuss than please share and you can share rational and irrational numbers type problems too.