Last visit was: 21 Jul 2024, 19:36 It is currently 21 Jul 2024, 19:36
Close
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
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Difficulty: 555-605 Level,   Probability,                                    
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642828 [248]
Given Kudos: 86716
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642828 [109]
Given Kudos: 86716
Send PM
avatar
Intern
Intern
Joined: 01 Aug 2006
Posts: 18
Own Kudos [?]: 144 [62]
Given Kudos: 0
Send PM
General Discussion
Manager
Manager
Joined: 20 Dec 2013
Posts: 183
Own Kudos [?]: 291 [8]
Given Kudos: 35
Location: India
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
6
Kudos
1
Bookmarks
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
User avatar
Manager
Manager
Joined: 11 Jan 2014
Posts: 78
Own Kudos [?]: 413 [2]
Given Kudos: 11
Concentration: Finance, Statistics
GMAT Date: 03-04-2014
GPA: 3.77
WE:Analyst (Retail Banking)
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
2
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.
Manager
Manager
Joined: 14 Jan 2013
Posts: 114
Own Kudos [?]: 1554 [13]
Given Kudos: 30
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE:Consulting (Consulting)
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
9
Kudos
3
Bookmarks
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
GMAT Club Legend
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6029
Own Kudos [?]: 13816 [3]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
3
Kudos
Expert Reply
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]

1x5
1x6
1x7
2x5
2x6
2x7

3x5
3x6
3x7
4x5
4x6
4x7


So probability = 8/12 = 2/3

I hope it helps!
GMAT Club Legend
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6029
Own Kudos [?]: 13816 [11]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
4
Kudos
7
Bookmarks
Expert Reply
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

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
User avatar
Jamboree GMAT Instructor
Joined: 15 Jul 2015
Status:GMAT Expert
Affiliations: Jamboree Education Pvt Ltd
Posts: 252
Own Kudos [?]: 658 [2]
Given Kudos: 1
Location: India
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
1
Kudos
1
Bookmarks
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.
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3036
Own Kudos [?]: 6609 [4]
Given Kudos: 1646
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
4
Kudos
Expert Reply
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.

Answer: D
User avatar
Intern
Intern
Joined: 16 Sep 2017
Posts: 7
Own Kudos [?]: 7 [0]
Given Kudos: 1
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
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
Tutor
Joined: 16 Oct 2010
Posts: 15126
Own Kudos [?]: 66769 [4]
Given Kudos: 436
Location: Pune, India
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
3
Kudos
1
Bookmarks
Expert Reply
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

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/
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30846 [0]
Given Kudos: 799
Location: Canada
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
Expert Reply
Top Contributor
Bunuel 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


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
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4918
Own Kudos [?]: 7808 [0]
Given Kudos: 221
Location: India
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
Top Contributor
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
Tutor
Joined: 05 Apr 2011
Status:Tutor - BrushMyQuant
Posts: 1803
Own Kudos [?]: 2145 [1]
Given Kudos: 100
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
1
Kudos
Expert Reply
Top Contributor
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!
Manager
Manager
Joined: 19 Nov 2022
Posts: 86
Own Kudos [?]: 19 [0]
Given Kudos: 1858
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
@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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642828 [1]
Given Kudos: 86716
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
1
Kudos
Expert Reply
SnorLax_7 wrote:
@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 ?


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.
Senior Manager
Senior Manager
Joined: 17 Sep 2016
Posts: 442
Own Kudos [?]: 85 [0]
Given Kudos: 147
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
JeffTargetTestPrep wrote:
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.

Answer: D


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.
Math Expert
Joined: 02 Sep 2009
Posts: 94441
Own Kudos [?]: 642828 [0]
Given Kudos: 86716
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
Expert Reply
zoezhuyan wrote:
JeffTargetTestPrep wrote:
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.

Answer: D


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.


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.
Senior Manager
Senior Manager
Joined: 17 Sep 2016
Posts: 442
Own Kudos [?]: 85 [0]
Given Kudos: 147
Send PM
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
Bunuel wrote:
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.


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
GMAT Club Bot
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink]
 1   2   
Moderator:
Math Expert
94441 posts