Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 20 Jul 2019, 11:23 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56304
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

13 00:00

Difficulty:   35% (medium)

Question Stats: 65% (01:28) correct 35% (01:50) wrong based on 313 sessions

### HideShow timer Statistics An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

_________________
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

Probability = Favorable outcome / Total Outcome

Total Outcomes = 10 x 10 = 100 (there will be 10 possibilities each time 'With replacement')

Favorable Outcome will be obtained when either both numbers are even or both numbers selected are odd. There are 5 even and 5 odd numbered balls for each selection of ball

1) Both Even: Therefore total outcomes = 5 x 5 = 25
2) Both Odd: Therefore total outcomes = 5 x 5 = 25

Total Favorable Outcomes = 25+25 = 50

Probability = 50/100 = 1/2 = 50%

_________________
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

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

2
Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

Alternate Method:

Probability = Favorable outcome / Total Outcome

Favorable Outcome will be obtained when EITHER both numbers are even OR both numbers selected are odd.

EVEN + EVEN = EVEN
OR
ODD + ODD = EVEN

Now,

Case 1: BOTH EVEN
Step 1) Probability of First ball being Even numbered = 5/10 = 1/2 (as there are 10 balls to pick from which have 5 even numbered and 5 odd numbered)
Step 2) Probability of Second ball being Even numbered = 5/10 = 1/2 (as there are 10 balls to pick from due to the condition 'WITH REPLACEMENT' which have 5 even numbered and 5 odd numbered)

Hence Probability of the sum being even as per Case-1 = (1/2) x (1/2) = 1/4

Case 2: BOTH ODD
Step 1) Probability of First ball being Odd numbered = 5/10 = 1/2 (as there are 10 balls to pick from which have 5 even numbered and 5 odd numbered)
Step 2) Probability of Second ball being Odd numbered = 5/10 = 1/2 (as there are 10 balls to pick from due to the condition 'WITH REPLACEMENT' which have 5 even numbered and 5 odd numbered)

Hence, Probability of the sum being even as per case-2 = (1/2) x (1/2) = 1/4

Total Probability = (1/4) + (1/4) = 1/2 = 50%

_________________
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

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

3
actually in this particular question we donot even require to calculate..
as we have equal number of odd and even number, the sum will be half the time even and half the time odd..
so ans 50%... C

also the numbers can be picked in four ways..
1)odd,odd
2)even,even
3)odd,even
4)even,odd
out of these 1 and 2 will give even number.. 3 and 4 will give odd number
so 2/4=50%..
_________________
Manager  Joined: 17 Apr 2013
Posts: 56
Location: United States
Concentration: Other, Finance
Schools: SDSU '16
GMAT 1: 660 Q47 V34 GPA: 2.76
WE: Analyst (Real Estate)
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

3
this is a straightforward C!
5 even 5 odds
So the first ball will have = 5/10 to be odd * 5/10 to have the second one odd
+
5/10 to be even * 5/10 to have the second one even
=
50%
_________________
Please +1 KUDO if my post helps. Thank you.
Intern  Joined: 03 Apr 2014
Posts: 1
GMAT 1: 550 Q48 V19 GPA: 4
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

1
Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

The Selection of balls:

Either both are even or odd

so in each case there is equal no - 5

Probability of drawing Even no one after one: 5/10 *5/10

Probability of drawing ODD no one after one: 5/10 *5/10

Either of them is possible, so Probility so that sum is either odd or even : ( 5/10 *5/10) + ( 5/10 *5/10) = 1/2 = 50%

Ans C
Math Expert V
Joined: 02 Sep 2009
Posts: 56304
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

1
Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Since there are 5 even numbered balls and 5 odd numbered balls, the probability of selecting either an even or an odd is 5/10= 1/2. In order for the sum of the numbers on the two balls to be even, either both numbers must be even or both numbers must be odd; one even and one odd number would add to an odd number. The probability of selecting two even numbers is:

1/2 * 1/2= 1/4

The probability of selecting two odd numbers is also:

1/2 * 1/2= 1/4

Adding both probabilities gives 1/4+1/4= 1/2 , or 50%, as the probability that the sum of the numbers will be even. Thus, the correct answer is (C).
_________________
Intern  S
Joined: 21 Jun 2015
Posts: 43
Location: India
Concentration: Finance, General Management
GMAT 1: 660 Q50 V30 GPA: 3.32
WE: Programming (Computer Software)
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

Probability = Favorable outcome / Total Outcome

Total Outcomes = 10 x 10 = 100 (there will be 10 possibilities each time 'With replacement')

Favorable Outcome will be obtained when either both numbers are even or both numbers selected are odd. There are 5 even and 5 odd numbered balls for each selection of ball

1) Both Even: Therefore total outcomes = 5 x 5 = 25
2) Both Odd: Therefore total outcomes = 5 x 5 = 25

Total Favorable Outcomes = 25+25 = 50

Probability = 50/100 = 1/2 = 50%
Manager  B
Joined: 11 Jun 2017
Posts: 68
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

Number of even values = 5
Number of odd values = 5

Total number of possibilities = 10*10 (chances of picking 10 at each go)

Even plus even and odd plus odd returns an even value, hence, total number of favorable outcomes = 5*5+5*5/ 100 = 1/2 = 50%
Math Expert V
Joined: 02 Sep 2009
Posts: 56304
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

30-sec approach:

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.

_________________
Intern  B
Joined: 10 Dec 2017
Posts: 21
GMAT 1: 680 Q48 V35 Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

ok my take
urn contains 10 balls of which 5 are even and 5 are odd
P = number of choices with the sum even / all the possible choices of picking 2 numbers
here we have replacement and order matters ====> number or all possible choices are 10x10
for a sum of 2 numbers to be even it can be either the sum of 2 even integers or the sum of 2 odd integers :
number of choices = 5x5 + 5x5 = 50 choices (remember we have 5 even and 5 odd integers in the urn)
P = 50/100 = 1/2 = 50%
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 937
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

1
Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

There are exactly 4 equiprobable events:

E1 = first ball odd , second ball odd
E2 = first ball odd, second ball even
E3 = first ball even, second ball odd
E4 = first ball even, second ball even

Exactly two of the 4 events mentioned above are favorable (E1 and E4), hence the correct answer is 50% (C).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14590
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Hi All,

We're told that an urn contains 10 balls, numbered from 1 to 10 and we're told that 2 balls are selected at random WITH replacement. We're asked for the probability that the SUM of the 2 numbers on the balls will be EVEN. This question can be solved in a couple of different ways, including as a straight-probability question.

To start, there are two ways to get a sum that is EVEN:

(Odd on the first) and (Odd on the second)
(Even on the first) and (Even on the second)

Since there are an equal number of Odd and Even numbers AND we are replacing the first ball before we pull the second, the probability of pulling one Odd or one Even is the same each time: 1/2

(Odd) and (Odd) = (1/2)(1/2) = 1/4
(Even) and (Even) = (1/2)(1/2) = 1/4
Total probability of ending with a sum that is Even on two balls is 1/4 + 1/4 = 2/4 = 1/2

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

# Rich Cohen

Co-Founder & GMAT Assassin Follow
Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Manager  S
Joined: 25 Mar 2018
Posts: 70
Location: India
Schools: ISB '21, IIMA , IIMB
GMAT 1: 650 Q50 V28 GPA: 4
WE: Analyst (Manufacturing)
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

First, Total number of outcomes by drawing 2 balls with replacement = 10 X 10 = 100

for even sum outcomes , either both should be even or both should be odd.
For both even we get 5C1 * 5C1 = 25 outcomes , for odd 5C1 * 5C1 = 25 outcomes , so there are 50 possible outcomes

So , the probability = 50/100 = 1/2 =50%
_________________
Please give me +1 kudos if my post helps you a little. It will help me unlock tests. Thanks
Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6967
Location: United States (CA)
Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  [#permalink]

### Show Tags

Bunuel wrote:
An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

Kudos for a correct solution.

The sum of the 2 numbers will be even if the 2 numbers chosen are both even or both odd. The probability of choosing 2 even numbers is 1/2 x 1/2 = 1/4. Likewise, the probability of choosing 2 odd numbers is 1/2 x 1/2 = 1/4. Therefore, the probability of choosing 2 numbers whose sum is even is 1/4 + 1/4 = 1/2.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select   [#permalink] 24 Mar 2019, 18:25
Display posts from previous: Sort by

# An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select  