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Re: W1 and W2 are two sets containing integers only. If an integer X is [#permalink]
amanvermagmat wrote:
W1 and W2 are two sets containing integers only. If an integer X is randomly selected from W1 and another integer Y is randomly selected from W2, what is the probability that (X*Y) is even?

(1) W1 = {-16, -4, 0, 6, 12}

(2) W2 = {-17, -15, -11, 19, 25}



lets take 2 smaller sets
1) W1= {-16}
2) W2 ={ -17,15}

{-16,-17 } {-16,15}
probability = 2/2=1

1) W1= {-16}
2) W2 ={ -17 ,15,11 }

{-16,-17 } {-16,15} {-16,11}

probability =3/3=1

So it seems if a set consists of all even elements , that set is sufficient .
Please can you share official solution .
Thanks .
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W1 and W2 are two sets containing integers only. If an integer X is [#permalink]
amanvermagmat wrote:
W1 and W2 are two sets containing integers only. If an integer X is randomly selected from W1 and another integer Y is randomly selected from W2, what is the probability that (X*Y) is even?

(1) W1 = {-16, -4, 0, 6, 12}

(2) W2 = {-17, -15, -11, 19, 25}


St 1: All are even integers so product with any other integer (even or odd) will always fetch an even integer. (Sufficient)

St 2: All are odd integers, therefore, if multiplied by even integer, result will be even, and if multiplied by odd integer result will be odd

So depends on W1's set of integers. Insufficient

Hence (A) Probability will always be 1 or 100% that the product of x*y is even.
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Re: W1 and W2 are two sets containing integers only. If an integer X is [#permalink]
A
As all elements in set w1 are even. The drawn integer x from set w1 will always make the product x*y even. In all scenarios, even without knowing the no of elements in set w2 or the elements in set w2, we can arrive at a sufficient conclusion that the product x*y is even.
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Re: W1 and W2 are two sets containing integers only. If an integer X is [#permalink]
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Re: W1 and W2 are two sets containing integers only. If an integer X is [#permalink]
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