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If x is to be chosen at random from the set {1,2,3} and y is to be

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energetics
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If x is to be chosen at random from the set {1,2,3} and y is to be [#permalink] New post   24 Jan 2019, 15:32
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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:
Show Spoiler
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] New post   24 Jan 2019, 21:51
Expert Reply
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:
Show Spoiler
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] New post   25 Jan 2019, 04:32
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:
Show Spoiler
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] New post   23 Aug 2023, 09:32
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?

Looking for X*Y = Even
Possibilities: EE, EO, OE (any combination that contains even will multiply to even)

Choose Even from Set X, Choose Even from Set Y = 1/3 x 2/4 = 1/6
Choose Even from Set X, Choose Odd from Set Y = 1/3 x 2/4 = 1/6
Choose Odd from Set X, Choose Even from Set Y = 2/3 x 2/4 = 1/3

1/6 + 1/6 + 1/3 = 4/6 or 2/3
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If x is to be chosen at random from the set {1,2,3} and y is to be [#permalink] New post   23 Aug 2023, 09:40
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?

Probability xy is even = 1 - Probability xy is odd = 1 - (2/3)(2/4)= 1 - (2/3)(1/2) = 1 - (1/3) = 2/3

Hence D
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If x is to be chosen at random from the set {1,2,3} and y is to be [#permalink]
23 Aug 2023, 09:40
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