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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

Problem Solving Question: 79 Category:Arithmetic; Algebra Probability; Concepts of sets Page: 72 Difficulty: 600

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
05 Feb 2014, 01:29

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Expert's post

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SOLUTION

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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

Probably the best way to solve would be to use 1 - P(opposite event) = 1 - P(odd) = 1 - P(odd)*P(odd) = 1 - 2/4*2/3 = 8/12 = 2/3.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
07 Feb 2014, 11:28

AKG1593 wrote:

Ans D

In order to make even nos. by multiplication, we should have even * odd, odd * even or even * even

Total even nos. possible by multiplying nos. from the 2 sets: (1*6); 2* any of the three from Set B; 3*6;& 4* any of the three from set B 1+3+1+3=8

Total possibilities=4 * 3=12

P(event)=8/12 or 2/3

I agree with you. But today, when I took the GMAT I encountered a similar question and my answer (according to that logic) wasn't there in the answer choices... So I assume it's not right.. I had x chosen at random from the numbers from 0 to 2 inclusive and y chosen at random from the numbers from 0 to 6 inclusive, what's the probability that x>y?

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
07 Feb 2014, 11:42

1

This post received KUDOS

x*y is even only when they're both even or one of them is odd. The probability of any single outcome is given by 1/3*1/4 = 1/12. Working through possible outcomes, we arrive at 8/12 -> 2/3, hence (D).

It took me around a minute and 40 seconds to solve this, I'm sure that a faster approach exists, waiting for others.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
07 Feb 2014, 11:56

3

This post received KUDOS

At least one of the numbers must be even. Both even: 1/2 * 1/3 = 1/6. At least one even: even from first set and odd from second set: 1/2 * 2/3 = 1/3 At least one even: odd from first set and even from second set: 1/2 * 1/3 = 1/6 Reqd prob = 1/6 + 1/3 + 1/6 = 2/3.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
08 Feb 2014, 03:08

Expert's post

SOLUTION

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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

Probably the best way to solve would be to use 1 - P(opposite event) = 1 - P(odd) = 1 - P(odd)*P(odd) = 1 - 2/4*2/3 = 8/12 = 2/3.

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
18 Feb 2014, 20:19

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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

The set {1, 2, 3, 4} contains 2 odd and 2 even numbers The set {5, 6, 7} contains 2 odd and 1 even numbers

Possible even xy = 2*2 + 2*2 = 8 and possible odd xy = 2*2 = 4

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
07 Apr 2014, 03:46

arunspanda wrote:

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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

I got it wrong the firs time, but 2nd time I used following method.

# of ways to select x = 4 # of ways to select y = 3 Total # of ways of getting xy = 4 * 3 = 12

If value of x is 1 then # of ways to get even value for xy = 1 (only 6 is possible value from set y) If value of x is 2 then # of ways to get even value for xy = 3 If value of x is 3 then # of ways to get even value for xy = 1 If value of x is 4 then # of ways to get even value for xy = 3

Total # of ways to get desired outcome = 1+3+1+3 = 8

Probability = desired outcome/ all possible outcomes = 8/12 = 2/3

Re: Probability- May i request you to help me in this one ? [#permalink]
27 Sep 2014, 06:46

1

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xy will be even when either x or y or both are even. Let's consider both x and y are old, so the product will be odd, not even. How many values can x take? There are only two odd values in the given set. Likewise, there are only two odd values possible for y. Since we want x and y to be odd, the total possible odd values for xy is 2*2=4. Total number of possible xy values is 4*3=12 (x can take any 4 values from the set and y can take any 3 values from the set). So probability of xy to be odd is 4/12=1/3. So probability of xy to be even is 1-(probability of being odd) = 1-1/3 = 2/3.

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

(A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6

Sol: Atleast one = total - none Now for xy to be even, Atleast one even should be there between x and y Atleast one even = total - no even In terms of probability, prob(atleast one even)= probab(total) - probab(no even I.e. Odd) = 1 - 2/4 * 2/3 = 2/3 As we can choose 2 odds out of 4 from set 1 and 2 odds out of 3 from set 2 _________________

Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
21 May 2015, 00:04

shreyas wrote:

At least one of the numbers must be even. Both even: 1/2 * 1/3 = 1/6. At least one even: even from first set and odd from second set: 1/2 * 2/3 = 1/3 At least one even: odd from first set and even from second set: 1/2 * 1/3 = 1/6 Reqd prob = 1/6 + 1/3 + 1/6 = 2/3.

In the case that both are even, shouldn't it matter which set we pick the even number from first?

Thus shouldn't the probability of picking 2 even numbers be: 1/2 * 1/3 * 2?

Are you considering only one case because the question stem asks about x*y only (and not y*x)? _________________

For all your days prepare, And meet them ever alike: When you are the anvil, bear— When you are the hammer, strike.

If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
21 May 2015, 00:30

1

This post received KUDOS

avgroh wrote:

shreyas wrote:

At least one of the numbers must be even. Both even: 1/2 * 1/3 = 1/6. At least one even: even from first set and odd from second set: 1/2 * 2/3 = 1/3 At least one even: odd from first set and even from second set: 1/2 * 1/3 = 1/6 Reqd prob = 1/6 + 1/3 + 1/6 = 2/3.

In the case that both are even, shouldn't it matter which set we pick the even number from first?

Thus shouldn't the probability of picking 2 even numbers be: 1/2 * 1/3 * 2?

Are you considering only one case because the question stem asks about x*y only (and not y*x)?

---short explanation-- question says probability that xy is even. Now xy = xy. So basically we need to find the probability of the product being even.

-- detailed explanation --for the case of both the numbers even the mathematical statement is simple-- one even number from set 1 AND one even number from set 2 (because we have to pic one number from each set). Order of picking is not relevant here. The question specifically says x is from first set and y is from the second set. The product has to be even and for a given value of (x,y), the product will be same whether you take product as xy or xy. The question is asking the probability of the "PRODUCT" being even.

Kudos .. If it helps! _________________

Impossibility is a relative concept!!

gmatclubot

If x is to be chosen at random from the set {1, 2, 3, 4} and
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21 May 2015, 00:30

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