It is currently 23 Mar 2018, 08:06

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 02 Sep 2015
Posts: 7
Schools: Haas
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

06 Oct 2015, 09:37
Gotcha! Thanks! I guess in this type of questions it's easiest to write out the possibilities?
Retired Moderator
Joined: 29 Oct 2013
Posts: 280
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

08 Dec 2015, 22:14
To avoid double counting it is better to just count one way. For instance,
1) You choose 2 0r 4 from set1 and any number from set2: probability = 1/2*1
2) You choose 1 or 3 from set1 and 6 from set2: Probability = 1/2*1/3= 1/6
Total Probability= 1/2+1/6= 8/12 = 2/3

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?

_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Intern
Status: About to write GMAT
Joined: 10 Mar 2015
Posts: 40
Location: India
GMAT 1: 700 Q49 V37
GPA: 3.9
WE: Web Development (Computer Software)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong:
If multiplication of numbers is even means at least one of the two numbers is even.

let a*b = even.
a is from first set. so probability of this being even: 2/4
b is from second set. so probability of this being even: 1/3
Accompanied by (or) so I added both.

I know this is wrong but am unable to figure out where I went wrong.
Kindly suggest a method in these lines.
Thanks
Intern
Status: About to write GMAT
Joined: 10 Mar 2015
Posts: 40
Location: India
GMAT 1: 700 Q49 V37
GPA: 3.9
WE: Web Development (Computer Software)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong:
If multiplication of numbers is even means at least one of the two numbers is even.

let a*b = even.
a is from first set. so probability of this being even: 2/4
b is from second set. so probability of this being even: 1/3
Accompanied by (or) so I added both.

I know this is wrong but am unable to figure out where I went wrong.
Kindly suggest a method in these lines.
Thanks
Manager
Joined: 09 Aug 2016
Posts: 68
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

22 Oct 2016, 11:25
The best way to solve this is to indetify that the product of being even can be obtained by the following cases:

x=E * y=O = E
x=O * y=E = E
x=E * y=E = E

The equation above might look quite silly but effectivly when I am saying x=E * y=O I mean that x is even and y is odd. As soon as you find this relationship you did the hard part of the question.

Then is all about replacing the prob/ties:
x=E * y=O in GMAT terms can be written as x=E AND y=0 means P(x=E) AND P(y=O) = 2/4 * 2/3 = 4/12 = 1/3 etc...

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

### Show Tags

21 Nov 2017, 12:11
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.
Can someone please clarify?

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?
Manager
Joined: 21 Oct 2017
Posts: 68
Concentration: Entrepreneurship, Technology
GMAT 1: 750 Q48 V44
WE: Project Management (Internet and New Media)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

02 Dec 2017, 18:14
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.
Can someone please clarify?

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

_________________

Please Press +1 Kudos if it helps!

October 9th, 2017: Diagnostic Exam - Admit Master (GoGMAT) - 640
November 11th, 2017: CAT 1 - Admit Master (GoGMAT) - 700
November 20th, 2017: CAT 2 - GMATPrep - 700 (Q: 47, V: 40)
November 25th, 2017: CAT 3 - Admit Master (GoGMAT) - 710 (Q: 48, V: 40)
November 27th, 2017: CAT 4 - GMATPrep - 720 (Q: 49, V: 40)

December 4th, 2017: GMAT Exam - 750 (Q: 48, V: 44, IR: 8, AWA: 6)

Math Expert
Joined: 02 Aug 2009
Posts: 5732
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

02 Dec 2017, 23:13
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.
Can someone please clarify?

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

Hi...

What you are doing here is " finding probability of picking an even number from atleast one of the set"
Say the set B were 5,6,8 so two even
Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1?

But we are looking for xy to be even..
This will include following cases..
1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6
3) odd from A and even from B
2/4*1/3=1/6

Total =1/3+1/6+1/6=2/3

Other easier way would be -
Subtract prob of odd from 1
Prob of odd .. odd from both = 2/4*2/3=1/3
So Ans=1 -1/3=2/3
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

GMAT online Tutor

Manager
Joined: 21 Oct 2017
Posts: 68
Concentration: Entrepreneurship, Technology
GMAT 1: 750 Q48 V44
WE: Project Management (Internet and New Media)
Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

03 Dec 2017, 10:35
chetan2u wrote:
Same question as above! A little confused why that doesn't work.

chetan2u can you please shed some light on this?

Thanks!

pablogutierrez24 wrote:
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer.
Can someone please clarify?

I reasoned: If any number in either set is even, then xy will be even.
So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)

Why does this result in the wrong probability?

Hi...

What you are doing here is " finding probability of picking an even number from atleast one of the set"
Say the set B were 5,6,8 so two even
Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1?

But we are looking for xy to be even..
This will include following cases..
1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6
3) odd from A and even from B
2/4*1/3=1/6

Total =1/3+1/6+1/6=2/3

Other easier way would be -
Subtract prob of odd from 1
Prob of odd .. odd from both = 2/4*2/3=1/3
So Ans=1 -1/3=2/3

Thanks for your response!

I guess I'm still a bit confused. For example, if a question was asking:

"A fair six-sided dice is rolled twice. What is the probability of getting at least once?"

Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 - (\frac{5}{6}*\frac{5}{6}) = \frac{11}{36}$$

Nevertheless, I could still say:
1. If I get 3 on my first roll I don't care what happen on the 2nd roll = $$\frac{1}{6}*\frac{6}{6}$$
2. If I don't get 3 on the 1st roll, I need 3 on the 2nd = $$\frac{5}{6}*\frac{1}{6}$$
3. Add the two results and get $$\frac{11}{36}$$

(Similarly, if the question was asking what is the probability that the sum of 2 rolls is 7, I would do $$1*\frac{1}{6}$$)

So here I went about calculating:
1. If I get even from the first set I don't care what the second set gets me = $$\frac{2}{4}*1$$
2. If I get an odd from the first set, I need an even from the second set =$$\frac{2}{4}*\frac{1}{3}$$

Now I understand this is wrong, and my first extinct would to subtract the probability of getting two odds. I just don't really understand why this is wrong lol. I'm thinking there is something about dependent / independent events I'm not completely getting...
_________________

Please Press +1 Kudos if it helps!

October 9th, 2017: Diagnostic Exam - Admit Master (GoGMAT) - 640
November 11th, 2017: CAT 1 - Admit Master (GoGMAT) - 700
November 20th, 2017: CAT 2 - GMATPrep - 700 (Q: 47, V: 40)
November 25th, 2017: CAT 3 - Admit Master (GoGMAT) - 710 (Q: 48, V: 40)
November 27th, 2017: CAT 4 - GMATPrep - 720 (Q: 49, V: 40)

December 4th, 2017: GMAT Exam - 750 (Q: 48, V: 44, IR: 8, AWA: 6)

VP
Joined: 22 May 2016
Posts: 1437
If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]

### Show Tags

11 Dec 2017, 16:50
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

chetan2u wrote:
Hi...

What you are doing here is " finding probability of picking an even number from atleast one of the set"
Say the set B were 5,6,8 so two even
Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1?

But we are looking for xy to be even..
This will include following cases..
1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6
3) odd from A and even from B
2/4*1/3=1/6

Total =1/3+1/6+1/6=2/3

Other easier way would be -
Subtract prob of odd from 1
Prob of odd .. odd from both = 2/4*2/3=1/3
So Ans=1 -1/3=2/3

Thanks for your response!

I guess I'm still a bit confused.
.
.
.
So here I went about calculating:
1. If I get even from the first set I don't care what the second set gets me = $$\frac{2}{4}*1$$
2. If I get an odd from the first set, I need an even from the second set =$$\frac{2}{4}*\frac{1}{3}$$

Now I understand this is wrong, and my first extinct would to subtract the probability of getting two odds. I just don't really understand why this is wrong lol. [I don't think it IS wrong.] I'm thinking there is something about dependent / independent events I'm not completely getting...

Hadrienlbb , I think the part in red is not accurate.
I am not sure where you think your mistake is, because if I finish your math, the answer is correct.

Here is the math from your steps, finished:
Quote:
1. If I get even from the first set I don't care what the second set gets me = $$\frac{2}{4}*1$$
$$= \frac{1}{2}$$

2. If I get an odd from the first set, I need an even from the second set =$$\frac{2}{4}*\frac{1}{3}$$
$$= \frac{2}{12} = \frac{1}{6}$$

Now add the results of your numbers 1 and 2:
$$(\frac{1}{2} + \frac{1}{6}) = (\frac{3}{6} + \frac{1}{6})= (\frac{4}{6}) = \frac{2}{3}$$

You just compressed chetan2u 's first two steps, which, with the third (and your second), he adds.

Here are his first two of three steps:

1) even from A and odd number from B...
2/4 * 2/3=1/3
2) even from A and even from B
2/4 *1/3=1/6

REWRITE:

(2/4 * 2/3) + (2/4 * 1/3) =

2/4 * (2/3 + 1/3) =

2/4 * (1) -- which is exactly what you calculated.
That's the result for picking an even number from set A.

Then add the result for picking an odd number from set A (which both you and he computed identically)

$$(2/4 * 1) + \frac{1}{6} =$$

$$(\frac{1}{2} + \frac{1}{6}) = \frac{6 + 2}{12}= \frac{8}{12} =\frac{2}{3}$$

What you did here is not the same as the person you were quoting . . . (and I had to delete all that because the machine hollers when there are too many quotes within quotes).

Maybe I am missing something. Now I am the confused one, I think.

It seems to me you got it right.
Why do you think there is a mistake?
_________________

At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

If x is to be chosen at random from the set {1, 2, 3, 4} and   [#permalink] 11 Dec 2017, 16:50

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.