GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Aug 2018, 16:38

### GMAT Club Daily Prep

#### 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

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

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 01 May 2007
Posts: 786
If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

Updated on: 17 Sep 2013, 07:23
2
1
16
00:00

Difficulty:

45% (medium)

Question Stats:

65% (00:53) correct 35% (01:01) wrong based on 748 sessions

### HideShow timer Statistics

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

Originally posted by jimmyjamesdonkey on 16 Jul 2007, 15:25.
Last edited by Bunuel on 17 Sep 2013, 07:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8195
Location: Pune, India
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

23 Jan 2014, 21:02
4
3
jimmyjamesdonkey 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

You can also use the sets formula:

P(xy is even) = P(x is even) + P(y is even) - P(Both x and y are even) (We subtract this to get rid of the double counting)
P(xy is even) = 2/4 + 1/3 - (2/4)(1/3) = 2/3

You can do many questions using the same concept. Check out this post for more on this: http://www.veritasprep.com/blog/2012/01 ... e-couples/
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to \$1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Director
Joined: 01 May 2007
Posts: 786

### Show Tags

16 Jul 2007, 16:51
11
5
OA is 2/3. Good job all. I actually got this wrong, but here is a great tip. I think everytime you see a probability problem, look at what it is asking, and then find the opposite of it.

For example, it is asking you to find the prob of xy being even...instead find the prob of it being ODD.

How is it odd? If x and y are both ODD.

Prob of x being odd: 1/2
prob of y being odd: 2/3
Prob of x and y being ODD: 1/2 * 2/3 = 1/3

Now 1 - 1/3 = 2/3..our answer.

I find that on all the OG problems dealing with prob, looking at it in reverse seems to bring you to a answer quicker. Anyone agree with that?
##### General Discussion
Manager
Joined: 24 Jun 2006
Posts: 124

### Show Tags

16 Jul 2007, 15:49
2
7
it should be D
even x even =1/2*1/3
evenx odd= 1/2* 2/3
odd*even= 1/2*1/3

P= 1/6+ 2/6+ 1/6= 2/3
Director
Joined: 12 Jun 2006
Posts: 524

### Show Tags

16 Jul 2007, 16:03
1
jimmyjamesdonkey wrote:
Q: 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

there are 12 different possibilities:
4C1 = 4
3C1 = 3
4(3) = 12

there's a 2/3 chance of set {5,6,7} giving us an odd # and there is a 1/2 chance of set {1,2,3,4} giving us an odd#. if only odd * odd = odd, there is a 1/3 = (2/3)*(1/2) chance of getting an odd product. therefore, there's a 8/12 ((1/3)*12 = 4) chance of getting an even product. 8/12 = 4/6 = 2/3
Intern
Joined: 17 Jun 2012
Posts: 1
Location: India
Schools: ISB '15, GMBA '15
GMAT Date: 03-06-2014
WE: Supply Chain Management (Transportation)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

23 Jan 2014, 14:01
I think we can use combinations too, to arrive at the answer.

The total number of combinations is 12 (4C1 for X * 3C1 for Y). The possibilities for having XY as even are:

1) when X is 1, Y has to be 6. That makes it 1C1*1C1
2) when X is 2, Y can take 5,6,7. This would be 1C1*3C1
3) when X is 3, Y can take only 6. So, this would mean 1C1*1C1
4) when X is 4, Y can take 5,6,7. That would be 1C1*3C1

So, 1C1*1C1 + 1C1*3C1 + 1C1*1C1 + 1C1*3C1 is divided by 4C1*3C1, i.e when 1 + 3 +1 + 3 is divided by 4*3 we get 2/3.

Please let me know if I am wrong.
SVP
Joined: 08 Jul 2010
Posts: 2139
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

23 Jul 2015, 06:47
2
2
jimmyjamesdonkey 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

Method-1
xy will be even when
1) x is even and y is odd, Probability of x even is (2/4) and Probability of y odd is (2/3), so Probability of Case(1) = (2/4)*(2/3) = (4/12)
2) x is Odd and y is Even, Probability of x Odd is (2/4) and Probability of y Even is (1/3), so Probability of Case(2) = (2/4)*(1/3) = (2/12)
3) x is even and y is Even, Probability of x even is (2/4) and Probability of y Even is (1/3), so Probability of Case(1) = (2/4)*(1/3) = (2/12)

Total Favorable Probability = (4/12)+(2/12)+(2/12) = (8/12) = 2/3

Method-2
Unfavorable case is when x and y are both odd,

Unfavorable Probability = (2/4)*(2/3) = 1/3

i.e. favorable probability = 1-(1/3) = 2/3

_________________

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

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Senior Manager
Joined: 28 Jun 2015
Posts: 295
Concentration: Finance
GPA: 3.5
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

23 Jul 2015, 07:30
1
Let A = {1,2,3,4}, B = {5,6,7}.

Total mappings from A->B = 4 * 3 = 12.

Number of mappings from A->B where A = {2,4} and B = {6}: 2 * 1 = 2 [even -> even]
Number of mappings from A->B where A = {1,3} and B = {6}: 2 * 1 = 2 [odd -> even]
Number of mappings from A->B where A = {2,4} and B = {5,7}: 2 * 2 = 4 [even -> odd]

Total favourable events = 2+2+4 = 8.

Probability = 8/12 = 2/3. Ans (D).
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2112
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

05 Oct 2015, 01:33
Set A={1, 2, 3, 4}
Set B={5, 6, 7}
For the product xy to be even , atleast one of the selected numbers should be even .
Method 1 = Prob(x is even and y is even) + Prob(x is even and y is odd) + Prob(x is odd and y is even)

Method 2=Prob(Favourable outcomes)= 1 - Prob(Unfavourable outcomes) =1- Prob(x is Odd and y is Odd)

The method of finding the unfavourable outcomes and subtracting the result from 1 will be quicker .

_________________

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful

Manager
Joined: 03 Jan 2017
Posts: 178
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

23 Mar 2017, 05:02
my approach was to start apply 1-xy (odd). As odd multiply can be only in case x and y are odd.

X set: 2/4
Y set: 2/3
2/4 and 2/3 = 4/12
1- 4/12=8/12=2/3

Another approach:
e*o=2/4*2/3
o*e=2/4*1/3
e*e=2/4*1/3
8/12 or 2/3
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2759
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

27 Mar 2017, 11:49
jimmyjamesdonkey 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

In order for xy to be even, at least one of the values of x and y needs to be even. We know that (1) even x even = even and (2) even x odd = even, and (3) odd x even = even.

Case 1. Both x and y are even. The probability that x is even is 2/4 = 1/2 , and the probability that y is even is ⅓; thus, the probability that x and y will both be even is 1/2 x 1/3 = 1/6

Case 2. x is even and y is odd. The probability that x is even is 1/2, and the probability that y is odd is ⅔; thus, the probability that x is even and y is odd is 1/2 x 2/3 = 2/6 = 1/3.

Case 3. x is odd and y is even. The probability that x is odd is ½, and the probability that y is even is ⅓; thus, the probability that x is odd and y is even is 1/2 x 1/3 = 1/6.

Thus, the total probability that the product xy will be even is 1/6 + 1/3 + 1/6 = 1/6 + 2/6 + 1/6 = 4/6 = 2/3.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Joined: 16 Sep 2017
Posts: 10
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

18 Sep 2017, 06:48
A noob question, if one set gives even does other set even matter as product would be even.

Probability to get even out of 1,2,3,4 - Is 2/4 or 1/2

Probability to get even out of 5,6,7 - Is 1/3

Shouldnt it be then

Pset1(even)*1(as other set doesn't matter-answer would be even) + Pset2(even)*1 =1/2+1/3.

Posted from my mobile device
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8195
Location: Pune, India
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

18 Sep 2017, 22:10
1
Anantz wrote:
A noob question, if one set gives even does other set even matter as product would be even.

Probability to get even out of 1,2,3,4 - Is 2/4 or 1/2

Probability to get even out of 5,6,7 - Is 1/3

Shouldnt it be then

Pset1(even)*1(as other set doesn't matter-answer would be even) + Pset2(even)*1 =1/2+1/3.

Posted from my mobile device

Here is the problem with this solution:

Pset1(even)*1 includes the probability that both elements are even.
Pset2(even)*1 also includes the probability that both elements are even.

So a case such as 2, 6 is counted twice. We need to remove once the probability that both are even.
Pset1(even)*Pset2(even) = (1/2)*(1/3) = 1/6

So to get the answer, 1/2 + 1/3 - 1/6 = 2/3

I have discussed the same thing here:
https://gmatclub.com/forum/if-x-is-to-b ... l#p1321459
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to \$1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 16 Sep 2017
Posts: 10
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and  [#permalink]

### Show Tags

18 Sep 2017, 22:26
2&6 and 6&2 getting in twice.

Got it thank you.

Posted from my mobile device
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and &nbs [#permalink] 18 Sep 2017, 22:26
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.