MacFauz wrote:
akhandamandala wrote:
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
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.
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.
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.