Set A consists of all the integers between 10 and 21, inclus

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.

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.

It is already given that y is being chosen from the list of prime numbers, and hence there will be a probability of 1 if we choose any number from y.
And for x, we have to find the probability of choosing a multiple of 3. Hence, the final probability is 4/12*1 = 1/3

After following all the discussion on this thread, I feel that considering only x is not correct. The question clearly states that y is the probability of selected a prime number from the set B. And then it asks for the probability of product xy being divisible by 3! Clearly we cannot assume that y will always be prime. We need to take it into consideration that we have picked y as a prime number from B and x as a factor of 3 from set A. The events x and y are definitely not mutually exclusive in this case as we need to find the probability for product xy . Also there can be cases where y if not prime could be a multiple of 3 and then xy would still be divisible by 3!

Please highlight if I'm overlooking something, But clearly the options and OA isnt correct.