From the given statements, we can conclude that y is prime. Therefore xy will be divisible by 3 if x is divisible by 3. Number of multiples of x between 10 and 21 inclusive = 4

Therefore probability = 4/12 = 1/3

Option B
_________________

If y has no factor z such that 1 < z < y, then y must be prime. Let's look at a few examples to
see why this is true:
6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime
15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime
3 has NO factor between 1 and 3: 3 IS prime
7 has NO factor between 1 and 7: 7 IS prime
Because it is selected from Set B, y is a prime number between 10 and 50, inclusive. The only prime
number that is divisible by 3 is 3, so y is definitely not divisible by 3.
Thus, xy is only divisible by 3 if x itself is divisible by 3. We can rephrase the question: “What is
the probability that a multiple of 3 will be chosen randomly from Set A?”
There are 21 – 10 + 1 = 12 terms in Set A. Of these, 4 terms (12, 15, 18, and 21) are divisible by 3

Thus, the probability that x is divisible by 3 is 4/12 = 1/3.
The correct answer is B. (Source : Manhattan GMAT Advanced Quant Guide)
_________________

we need to discuss a little bit here.

I agree that y is prime. However y is also chosen, so the probability for a prime number chosen is 13/41. And the probability for x divisible by 3 is 4/12
So the result is (13/41)*(4/12).

Please correct me if I'm wrong, thanks

We do not have to worry about choosing y. We are already given that y is prime. So no matter what number we choose as y, we still have to only find out the probability for choosing x. If otherwise, the question had asked for the number of ways in which xy would be divisible by 3, then the number of ways in which y could be selected would have to have been considered.

Would it have been complex if it were given that B consists of all prime numbers from 1 to 50, inclusive?
Will it be \(1/3 + 1/15\) then?
_________________

Is that not what the question is already effectively asking??

The question says that B consists of all integers from 10 to 50, inclusive.
_________________

True.. But the options to choose Y is essentially a set of prime numbers between 10 and 50...

I agree that indirectly it is stating that B consists of prime numbers from 10 to 50, but my query was rather different.
Suppose, if it were given that B consists of all prime numbers from 1 to 50 rather than 10 to 50, then what will be the answer.
I hope, I am being clear this time.
_________________

Oh.. Sorry. I read that as 10 to 50 again... I'm not very sure. But let me give it a try

edit:

If it is 1 to 50,

For required outcome : y is 3 and x is any number or y is not 3 and x is a multiple of 3

\(\frac{1}{15} * 1 + \frac{14}{15} * \frac{4}{12}\)

\(= \frac{12}{15*12} + \frac{14*4}{15*12}\)

\(= \frac{68}{180} = \frac{17}{45}\)

A very ugly answer and lots of calculator work.. There should be an easier more correct way..

Thanks for your reply. I would like to defend my idea:

"Set B consists of all the integers between 10 and 50, inclusive" means set B will contain prime and non-prime numbers
"If x is a number chosen randomly from Set A" means x is chosen randomly from set A without limitation condition
"y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y" means y is also chosen like x but with condition.
So y is not a given number, y must be chosen from set B and satisfy the condition "has no factor z such that 1 < z < y".

Now take a look from another view:
"Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, what is the probability that the product xy is divisible by 3?"
The condition for y in this version is removed, so what is your idea in this case ???

In such a case, there are 13 ways in which y can be selected to be a multiple of 3 and 28 ways in which y can be selected to not be a multiple of 3.

\(\frac{13}{41} * 1 + \frac{28}{41}* \frac{4}{12}\)

\(= \frac{67}{123}\)

i believe the fact the y is a prime number greater than 10 given in the original question is significant because it clearly says that y CANNOT be a multiple of 3.

That's very good. We step by step go to the mutual agreement that "y is a number chosen randomly from Set B" means there are various ways in which y is chosen; so y may be a prime or may be a non-prime number. What is the probability that y will be a prime number?

MacFauz why have you neglected the 3 in the other one? Can't we take any number just as it was done earlier?
_________________

@akhandamandala : I believe that in the original question, it is given that "y" is a prime number. So, the probability that y will be prime is 100%.

@Marcab : I have already considered a situation where y is 3 and x is a multiple of 3 in the first scenario. Hence, in the second scenario, I am considering only a situation where y is not a multiple of 3 and x is a multiple of 3. If in the second case, "y" is not 3 and "x" is also not a multiple of 3, then the product would not be divisible by 3.

Set B does not contains prime numbers only, that's the set of ALL integers from 10 to 50, and there are 11 (not 13 as aforementioned) prime numbers in the range [10, 50]. In the other hand, y is chosen RANDOMLY from the set B, and y is prime; so P(y is prime) = 1 is not persuasive.

I posted the original quiz on facebook (without the multiple choices), there's a reply with the idea that the probability for y is prime is <1. So far my idea does not stand alone.
Why don't you make a survey in which you post this original quiz (omit the multiple choices) on your facebook or blog?

I think this question is very controversial. There would be significant amount of people understand that y is chosen randomly in order to satisfy the condition y is prime, not that y is given as a prime.

y has no factor z such that 1 < z < y this means that y is a prime and it has to lie between 10 and 50 inclusive. Now, since the product of xy is divisible by 3 so that means that we have to look for multiples of 3 in set A since primes from set B are all greater than 3. Set B has 12 elements and out of which 4 are multiples of 3 so probability should be 4/12 = 1/3.

Consider giving kudos if my post helped.

I'm little bit confused here.
The way GyanOne provided is pretty logical and I like it but I have a conundrum with more choosey approach.

In Set A we have 12 integers, these ones divisble by 3 - 12, 15,18, 21 - 4 numbers
In set B we have 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 - 11 prime numbers

prob of xy divisible by 3 = (xy's divisble by 3)/(total possible xy's)

xy's divisible by 3= 11x4=44
total possible xy's= 11x12-4=128, I deduct 4 to avoid double counting of 11, 13, 17,19 as these appear in both sets

so, I have 44/128=11/32, slightly more that 1/3.

please, tell me where am I wrong here.