by: SUZYC
17 mins
Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:28) correct
27%
(01:31)
wrong
based on 7531
sessions
History
Date
Time
Result
Not Attempted Yet
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
_________________
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
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.
Answer: D.
_________________
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.
Answer: D.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
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.
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.
3 Scenario's
E * E = 2/4 * 1/3 = 1/6 ----- 1
E * O= 2/4 * 2/3 = 4/12-----2
O* E = 2/4 *1/3 = 2/12------3
Add 1+2+3 ==> 2/3 ----------D
E * E = 2/4 * 1/3 = 1/6 ----- 1
E * O= 2/4 * 2/3 = 4/12-----2
O* E = 2/4 *1/3 = 2/12------3
Add 1+2+3 ==> 2/3 ----------D
General Discussion
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
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
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.
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.
BrainLab
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!
_________________
GMATinsight (Tutor & Admission Consultant) | Contact for a FREE trial session
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
jimmyjamesdonkey
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
Answer: option D
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
Answer: option D
_________________
GMATinsight (Tutor & Admission Consultant) | Contact for a FREE trial session
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
AryamaDuttaSaikia
Jamboree GMAT Instructor
Joined: 15 Jul 2015
Last visit: 06 Dec 2019
Posts: 250
Status:GMAT Expert
Affiliations: Jamboree Education Pvt Ltd
Location: India
Posts: 250
If you write the cases, they are {(2,5) (2,6) (2,7)}, {(4,5),(4,6),(4,7)} & {(6,1),(6,2),(6,3),(6,4)}
As you can observe, there are two common cases out of total options of 12. So, probability = 2/12=1/6 needs to subtracted from the addition .
So, 1/2+1/3-1/6 = 4/6 =2/3
Hope this helps.
_________________
As you can observe, there are two common cases out of total options of 12. So, probability = 2/12=1/6 needs to subtracted from the addition .
So, 1/2+1/3-1/6 = 4/6 =2/3
Hope this helps.
_________________
Aryama Dutta Saikia
Jamboree Education Pvt. Ltd.
Jamboree Education Pvt. Ltd.
JeffTargetTestPrep
Expert
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
jimmyjamesdonkey
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.
Answer: D
_________________
Jeffrey Miller | Head of GMAT Instruction |
[email protected]
See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews
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
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
Anantz
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
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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... e-couples/
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Posts: 6,727
Bunuel
One approach is to recognize that P(xy is even) = 1 - P(xy is not even)
In other words, P(xy is even) = 1 - P(xy is odd)
Aside: This is a useful approach since there's only one way that xy can be odd. Both x and y must be odd for their product to be odd.
Conversely, there are 3 different cases to consider for xy to be even: 1) x and y are both even. 2) x is odd and y is even. 3) x is even and y is odd.
P(xy is odd) = P(x is odd AND y is odd)
= P(x is odd) x P(y is odd)
= 2/4 x 2/3
= 1/3
So, P(xy is even) = 1 - 1/3
= 2/3
Answer: D
Cheers,
Brent
Posts: 4,849
Solution:
Probability of choosing an odd number from set X = 2/4
Probability of choosing an odd number from set Y = 2/3
The possibilities for choosing a number from set X that is Odd and another from set Y that is also odd
= 2/4 * 2/3
= 1/3
Now, 1- 1/3 = 2/3 is the probability of choosing two numbers which can be even, odd or even, even or odd, even implying the product xy will always be even.
(option d)
Devmitra Sen
GMAT SME
Probability of choosing an odd number from set X = 2/4
Probability of choosing an odd number from set Y = 2/3
The possibilities for choosing a number from set X that is Odd and another from set Y that is also odd
= 2/4 * 2/3
= 1/3
Now, 1- 1/3 = 2/3 is the probability of choosing two numbers which can be even, odd or even, even or odd, even implying the product xy will always be even.
(option d)
Devmitra Sen
GMAT SME
Given that 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} and we need to find what is the probability that xy will be even
We know that Total Probability is always equal to 1
=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)
To get xy as odd we need to have all of them as odd.
=> P(xy = Odd) = P(x = odd) * P(y = odd) = \(\frac{2}{4}\) * \(\frac{2}{3}\)
(Because for x we have two choices (1,3) to get odd out of 4, for y we have two choices (5,7) to get odd out of 3)
= \(\frac{1}{3}\)
=> P(Even) = 1 - P(Odd) = 1 - \(\frac{1}{3}\) = \(\frac{3 - 1}{3}\) = \(\frac{2}{3}\)
So, Answer will be D.
Hope it helps!
_________________
We know that Total Probability is always equal to 1
=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)
To get xy as odd we need to have all of them as odd.
=> P(xy = Odd) = P(x = odd) * P(y = odd) = \(\frac{2}{4}\) * \(\frac{2}{3}\)
(Because for x we have two choices (1,3) to get odd out of 4, for y we have two choices (5,7) to get odd out of 3)
= \(\frac{1}{3}\)
=> P(Even) = 1 - P(Odd) = 1 - \(\frac{1}{3}\) = \(\frac{3 - 1}{3}\) = \(\frac{2}{3}\)
So, Answer will be D.
Hope it helps!
_________________
How to Solve :
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
@bunuel/@chetan2u/@KarishmaB
I have understood the logic of the question, But how you have arrived at the conclusion that 'xy' is a product and not a random number from Set A and Set B. (for eg. 15, 16, 17) because in that case, probabality would have been 1/3.
Kindly can you help ?
I have understood the logic of the question, But how you have arrived at the conclusion that 'xy' is a product and not a random number from Set A and Set B. (for eg. 15, 16, 17) because in that case, probabality would have been 1/3.
Kindly can you help ?
SnorLax_7
xy there means x*y, x multiplied by y not a two-digit integer xy, recall that multiplication sign is often omitted. If xy were a two-digit number, it would have been mentioned explicitly.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
JeffTargetTestPrep
hi JeffTargetTestPrep
I did not consider case 1, because I thought case 2 &3 include case 1, leading me to pick up E
please clarify ?
thanks in advance.
zoezhuyan
There are four complementary, and thus mutually exclusive, scenarios when choosing two numbers: EE, EO, OE, and OO.
- EE: Probability is 1/6
- EO: Probability is 1/3
- OE: Probability is 1/6
- OO: Probability is 1/3
The total probability adds up to 1 (1/6 + 1/3 + 1/6 + 1/3 = 1). The first three scenarios (EE, EO, OE) result in an even product xy. Therefore, scenarios 2 and 3 do not include scenario 1.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
Bunuel
hi Bunuel
say case 2, x=even, y = whatever
then x is selected from 2 or 4,
y selected from 5,6,7 -- so it is highly possible to picked up x=2, y=6
say case 1, both x and y are even
so it is possible x=2,y=6,
you seem case 1 is already included in case 2
that's why I didn't consider case1,
genuinely need your further clarify.
thanks a lot
Moderator:
Announcements
How to Tell If Your GMAT Prep Resources Are Actually Helping You
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Tuck at Dartmouth
AI at Dartmouth Tuck: How the School Is Preparing Leaders for the Age of Artificial Intelligence
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Top 5 GMAT Debrief Videos
Success stories and strategies from high-scoring candidates.