Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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]

Show Tags

07 Feb 2014, 12:28

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]

Show Tags

07 Feb 2014, 12:42

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]

Show Tags

07 Feb 2014, 12:56

5

This post received KUDOS

2

This post was BOOKMARKED

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.

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]

Show Tags

18 Feb 2014, 21:19

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]

Show Tags

07 Apr 2014, 04:46

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]

Show Tags

27 Sep 2014, 07:46

1

This post received KUDOS

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]

Show Tags

21 May 2015, 01:04

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)?

If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

Show Tags

21 May 2015, 01:30

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.

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

Show Tags

30 Jun 2015, 10:21

Why do we count here only possibilities from Set 1 to Set 2 and not also vice versa as with this example

Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?

Here are the rolls that work: 2 - 6 3 - 5 4 - 4 5 - 3 6 - 2 That’s it; there are 5 combinations that work. Therefore the probability of a sum of 8 is 5/36.
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Why do we count here only possibilities from Set 1 to Set 2 and not also vice versa as with this example

Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that the sum of the rolls is 8?

Here are the rolls that work: 2 - 6 3 - 5 4 - 4 5 - 3 6 - 2 That’s it; there are 5 combinations that work. Therefore the probability of a sum of 8 is 5/36.

The example that you have quoted here is the example of arrangements

When we say total Outcomes of two rolls = 6 x 6 = 36 [It includes all arrangements like 1-6 and 6-1 separately]

However, when we calculate the Products the the arrangement doesn't matter 2*3 is same as 3*2 because the result is 6 which is identical

When we say total Outcomes in this case = 4 * 3 = 12 [It considers all possible multiplication just once]

1x5 1x6 1x7 2x5 2x6 2x7 3x5 3x6 3x7 4x5 4x6 4x7

So probability = 8/12 = 2/3

I hope it helps!
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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

Show Tags

27 Sep 2015, 13:07

Hey all, I'm having trouble understanding why my approach doesn't work and I'd really appreciate your help! I picked E because for xy to be even, you need to pick an even number in first set (and you don't care about what you pick in second set), or an even number in second set (and you don't care what you pick in first set). So then the probability of even in first set is 1/2 and second set is 1/3. Why can't I just add the probabilities together to have 5/6? What have I included in the probability to make it 1/6 bigger than the right answer?

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...