GMAT Changed on April 16th - Read about the latest changes here

It is currently 26 Apr 2018, 08:09

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

There are 10 balls numbered 1-10 and two balls are picked WITH and WIT

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

Hide Tags

1 KUDOS received
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 651
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 19 Dec 2010, 12:31
1
This post received
KUDOS
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:14) wrong based on 5 sessions

HideShow timer Statistics

There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

[Reveal] Spoiler:
Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?

_________________

Consider kudos, they are good for health

Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 145
Location: Toronto
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 19 Dec 2010, 12:37
mainhoon wrote:
There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?


You're not wrong!

To get an even sum, you must add two odds or two evens; to get an odd sum, you add an odd and an even.

So, in the "without replacement" case, you have a better chance to get an odd sum.

Let's review the 4 cases:

odd then even: 5/10 * 5/9 = 25/90
even then odd: 5/10 * 5/9 = 25/90

odd then odd: 5/10 * 4/9 = 20/90
even then even: 5/10 * 4/9 = 20/90

As you can see, there's a 50/90 chance of getting an odd sum and only a 40/90 chance of getting an even sum.

In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 651
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 19 Dec 2010, 12:57
Ah thanks for the quick reply. I guess I was thinking conceptually why should odd be favored over even. In the with replacement case so it is 50-50 to get an odd or even sum, but in the without replacement case, odds for odd are higher than even.. Well a followup.

I solved the question with the combination formula for the even-without-replacement case. Can you show how to solve this with the combination formula with the odd-without-replacement case?

2 of 10 balls = 10C2 = 45
Odd Sum: Odd-Even or Even-Odd: 5C1 x 5C1 + 5C1 x 5C1 = 50? [This should really be 25]

What is wrong here?
_________________

Consider kudos, they are good for health

Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 145
Location: Toronto
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 19 Dec 2010, 21:36
mainhoon wrote:
Ah thanks for the quick reply. I guess I was thinking conceptually why should odd be favored over even. In the with replacement case so it is 50-50 to get an odd or even sum, but in the without replacement case, odds for odd are higher than even.. Well a followup.

I solved the question with the combination formula for the even-without-replacement case. Can you show how to solve this with the combination formula with the odd-without-replacement case?

2 of 10 balls = 10C2 = 45
Odd Sum: Odd-Even or Even-Odd: 5C1 x 5C1 + 5C1 x 5C1 = 50? [This should really be 25]

What is wrong here?


Hi,

since it's a combinations problem, order doesn't matter - there are 5 odd balls and we want one, so there's 5C1 ways to choose; similarly, there are 5 even balls and we want to choose one, so there's 5c1 ways to choose. We're choosing one odd AND one even, so we multiply:

5C1 * 5C1 = 5 * 5 = 25

Your mistake was considering odd then even and even then odd as two different possibilities - that's why you got double the actual answer.
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44650
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 20 Dec 2010, 03:26
mainhoon wrote:
There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?


WITHOUT REPLACEMENT CASE



Combinatorial approach:
Odd sum: \(P(eo)+P(oe)=\frac{C^1_5*C^1_5+C^1_5*C^1_5}{C^1_{10}*C^1_{9}}=\frac{50}{90}\) or which is the same \(P(odd \ sum)=2*\frac{C^1_5*C^1_5}{C^1_{10}*C^1_{9}}=\frac{50}{90}\);

Note here that \(C^2_{10}\) won't give you the # of total outcomes (denominator) as C here stands for combinations and by definition it gives the # of selections of 2 different objects out of 10 when order of the selection doesn't matter, but total # of outcomes here is 10*9=90 (1-2, 1-3, 1-4, ..., 2-1, ..., 10-9) so for total # of outcomes order matters as for example 2-4 is different outcome than 4-2.

Here you can solve the question so that to have \(C^2_{10}\) in the denominator. Though in this case \(C^2_{10}\) will represent all possible pairs of 2 numbers out of 10, some will give odd sum and other even. Thus nominator also should be based on the same approach: \(C^1_5*C^1_5\) will give all possible pairs of odd and even numbers (as we are interested in a selection of odd-even pair then the order doesn't matter), so \(P=\frac{C^1_5*C^1_5}{C^2_{10}}=\frac{25}{45}=\frac{50}{90}\).

Even sum: \(P(ee)+P(oo)=\frac{C^1_5*C^1_4+C^1_5*C^1_4}{C^1_{10}*C^1_{9}}=\frac{40}{90}\);

Here you can also solve the question so that to have \(C^2_{10}\) in the denominator. Again \(C^2_{10}\) will represent all possible pairs of 2 numbers out of 10, some will give odd sum and other even. Thus nominator also should be based on the same approach: \(C^2_5\) will give all possible pairs of 2 even numbers out of 5 and \(C^2_5\) will give all possible pairs of 2 odd numbers out of 5 (as both will give even sum), so \(P=\frac{C^2_5+C^2_5}{C^2_{10}}=\frac{20}{45}=\frac{40}{90}\).

Probability approach:
Odd sum: \(P(eo)+P(oe)=\frac{5}{10}*\frac{5}{9}+\frac{5}{10}*\frac{5}{9}=\frac{50}{90}\) or which is the same \(P(odd \ sum)=2*\frac{5}{10}*\frac{5}{9}=\frac{50}{90}\), multiplying by 2 as odd-even can occur in two ways OE and EO;

Even sum: \(P(ee)+P(oo)=\frac{5}{10}*\frac{4}{9}+\frac{5}{10}*\frac{4}{9}=\frac{40}{90}\);


WITH REPLACEMENT CASE


No matter what ball you pick for the first one, you'll have 1/2 chances to make the sum even with the second ball and the same 1/2 chances to make the sum odd.

Similar problem: http://gmatclub.com/forum/probability-q ... en#p643043

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 69
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 21 Dec 2010, 07:24
skovinsky wrote:
mainhoon wrote:
In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.


Hi
I don´t understand what you want to say here.
For the "with replacement" scenario I simply have 2 different cases:
when first is odd, second is even \(\frac{5}{10}*\frac{5}{10}\)
when first is even, second is odd \(\frac{5}{10}*\frac{5}{10}\)

That is \(\frac{5}{10}*\frac{5}{10}*2=\frac{1}{2}\)
Is that what you mean?
Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 145
Location: Toronto
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 21 Dec 2010, 19:29
medanova wrote:
skovinsky wrote:
mainhoon wrote:
In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.


Hi
I don´t understand what you want to say here.
For the "with replacement" scenario I simply have 2 different cases:
when first is odd, second is even \(\frac{5}{10}*\frac{5}{10}\)
when first is even, second is odd \(\frac{5}{10}*\frac{5}{10}\)

That is \(\frac{5}{10}*\frac{5}{10}*2=\frac{1}{2}\)
Is that what you mean?


Hi,

I outlined 4 cases in my post (o/o, o/e, e/o, e/e). In the "with replacement" scenario, each is equally likely (i.e. has a 5/10 * 5/10 = 25/100 chance to happen). Since 2 of the 4 cases match what we want, there's a 25/100 + 25/100 = 50/100 probability of getting what we desire.
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 6648
Premium Member
Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT [#permalink]

Show Tags

New post 14 Dec 2017, 05:49
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT   [#permalink] 14 Dec 2017, 05:49
Display posts from previous: Sort by

There are 10 balls numbered 1-10 and two balls are picked WITH and WIT

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


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.