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Senior Manager  P
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If x is to be chosen at random from the set {1,2,3} and y is to be  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 65% (01:17) correct 35% (02:37) wrong based on 32 sessions

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If x is to be chosen at random from the set {1,2,3} and y is to be chosen at random from the set {4,5,6,7}, what is the probability that xy will be even?

(A) $$\frac{1}{6}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) $$\frac{5}{6}$$

Explanation:
For xy to be even, either x, y, or both must be even. The only way xy will not be even is if both x and y are odd. Since the latter is only one probability, while the former is three, it’s most efficient to solve for the latter– the opposite of what the question is asking for. Just remember to subtract the result from 1 before selecting an answer.

The probability that both numbers are odd relies on the probabilities that each individual number is odd. In the set that x is chosen from, there are 3 numbers, 2 of which are odd. Thus, the probability that x is odd is 2/3
3 . In the set that y is chosen from, there are 4 numbers, 2 of which are odd. Thus, the probability that y is odd is
1/2. The probability that both are odd, then, is: p = (2/3)*(1/2) = 1/3

Remember that 1/3 is the opposite of what we’re solving for; subtract that from 1: p = 1 − 1/3 = 2/3 , choice (D).
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Re: If x is to be chosen at random from the set {1,2,3} and y is to be  [#permalink]

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energetics wrote:
If x is to be chosen at random from the set {1,2,3} and y is to be chosen at random from the set {4,5,6,7}, what is the probability that xy will be even?

(A) $$\frac{1}{6}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) $$\frac{5}{6}$$

Explanation:
For xy to be even, either x, y, or both must be even. The only way xy will not be even is if both x and y are odd. Since the latter is only one probability, while the former is three, it’s most efficient to solve for the latter– the opposite of what the question is asking for. Just remember to subtract the result from 1 before selecting an answer.

The probability that both numbers are odd relies on the probabilities that each individual number is odd. In the set that x is chosen from, there are 3 numbers, 2 of which are odd. Thus, the probability that x is odd is 2/3
3 . In the set that y is chosen from, there are 4 numbers, 2 of which are odd. Thus, the probability that y is odd is
1/2. The probability that both are odd, then, is: p = (2/3)*(1/2) = 1/3

Remember that 1/3 is the opposite of what we’re solving for; subtract that from 1: p = 1 − 1/3 = 2/3 , choice (D).

This is a copy of the following OG question: https://gmatclub.com/forum/if-x-is-to-b ... 67089.html
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Re: If x is to be chosen at random from the set {1,2,3} and y is to be  [#permalink]

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P odd no xy ;

2/3 * 2/4 = 1/3
even 1-1/3 = 2/3 IMO D

energetics wrote:
If x is to be chosen at random from the set {1,2,3} and y is to be chosen at random from the set {4,5,6,7}, what is the probability that xy will be even?

(A) $$\frac{1}{6}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) $$\frac{5}{6}$$

Explanation:
For xy to be even, either x, y, or both must be even. The only way xy will not be even is if both x and y are odd. Since the latter is only one probability, while the former is three, it’s most efficient to solve for the latter– the opposite of what the question is asking for. Just remember to subtract the result from 1 before selecting an answer.

The probability that both numbers are odd relies on the probabilities that each individual number is odd. In the set that x is chosen from, there are 3 numbers, 2 of which are odd. Thus, the probability that x is odd is 2/3
3 . In the set that y is chosen from, there are 4 numbers, 2 of which are odd. Thus, the probability that y is odd is
1/2. The probability that both are odd, then, is: p = (2/3)*(1/2) = 1/3

Remember that 1/3 is the opposite of what we’re solving for; subtract that from 1: p = 1 − 1/3 = 2/3 , choice (D). Re: If x is to be chosen at random from the set {1,2,3} and y is to be   [#permalink] 25 Jan 2019, 04:32
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