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
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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

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

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?

@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.

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.

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.