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Probability

by PGMAT » Thu Jul 05, 2012 7:05 pm
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, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?

1/4
1/3
1/2
2/3
3/4

[spoiler]OA is B.OE - the probability of choosing a multiple of 3 from set A is 1/3. My question, shouldn't we consider the probability of choosing a prime number from set B and then take a product of both probabilities? Example, y=17 and x=12?[/spoiler]

Thanks.
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by Anurag@Gurome » Thu Jul 05, 2012 7:47 pm
PGMAT wrote: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, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?
Note that from the definition y is nothing but a prime integer.

Set A has (21 - 10 + 1) = 12 integers
Set B has (50 - 10 + 1) = 41 integers

Hence, total number of possible values of xy = 12*41

Now, multiples of 3 in set A are: 12, 15, 18, and 21 --> 4 integers
And, prime numbers in set B are: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47 --> 11 integers

Hence, total number of possible values of xy such that xy is a multiple of 3 = 4*11

Hence, required probability = (4*11)/(12*41) = 11/(3*41) = 11/123

Note : The answer would have been B if set B had only the prime integers between 10 and 50.
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