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.

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

Show Tags

05 Feb 2014, 02:29

1

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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]

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

4

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]

Show Tags

08 Feb 2014, 04:08

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]

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 !

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

Show Tags

30 Jun 2015, 10:34

Expert's post

BrainLab wrote:

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]

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...