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anonymous
 5 years ago
if two numbers x and y are chosen at random from the set of first 30 natural numbers. the chance that x^2y^2 is divisible by 3 is
anonymous
 5 years ago
if two numbers x and y are chosen at random from the set of first 30 natural numbers. the chance that x^2y^2 is divisible by 3 is

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ChaChing, awards in the bank.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2y^2= (x+y)(xy) so we need to check what is the chance that x+y or xy is divisible of 3 if x is divisible of 3 (1/3 chance) than there is 1/3 chance that x+y will be divisible and in that case both are (so xy) too if x is not divisible of 3 (2/3)than if y is not divisible of 3 (2/3) chance than either x+y or xy will be factor of 3. If y is divisible of 3 than neither x+y and xy will be divisible. So the answer is 1/3*1/3+2/3*2/3=5/9 I am completely not sure if this is correct :) But I thought about it for some time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no the answer is 47/87

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you see where my logic failed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"if x is divisible of 3 (1/3 chance) than there is 1/3 chance that x+y will be divisible and in that case both are (so xy) too" why do you say 1/3 chance?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that x+3 and xy are divisible of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how does dividing x by 3 has onethird chance?? m not gettin it..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/3*9/29+2/3*19/29= 47/87

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my logic was good but I did not see that it is without replacement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/3 chance that x is divisible of 3 than x+y or xy only divisible of 3 if y is divisible of 3, that is 9/29 2/3 chance that x is not divisible of 3 than if y is not divisible of 3 than either x+y or xy will be. and that is 19/29
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