Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56304

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
16 Feb 2015, 07:00
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
WE: Education (Education)

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
16 Feb 2015, 07:15
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% Answer: Option
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2960
Location: India
GMAT: INSIGHT
WE: Education (Education)

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
16 Feb 2015, 07:24
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 = EVENNow, 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 Case1 = (1/2) x (1/2) = 1/4Case 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 case2 = (1/2) x (1/2) = 1/4Total Probability = (1/4) + (1/4) = 1/2 = 50%Answer: Option
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Math Expert
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
17 Feb 2015, 07:56
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
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
17 Feb 2015, 17:54
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
GPA: 4

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
17 Feb 2015, 19:18
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
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
22 Feb 2015, 11:27
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 SOLUTIONCorrect Answer: C 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
Joined: 21 Jun 2015
Posts: 43
Location: India
Concentration: Finance, General Management
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
22 Sep 2017, 00:10
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
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
22 Sep 2017, 08:58
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
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
14 Dec 2017, 05:47
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. 30sec 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. Answer: C.
_________________



Intern
Joined: 10 Dec 2017
Posts: 21

Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
14 Sep 2018, 01:35
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
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
27 Nov 2018, 19:26
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 highlevel "quant" preparation starts here: https://gmath.net



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14590
Location: United States (CA)

Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
27 Nov 2018, 22:39
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 straightprobability 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 Final Answer: 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
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



Manager
Joined: 25 Mar 2018
Posts: 70
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Analyst (Manufacturing)

Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select
[#permalink]
Show Tags
27 Nov 2018, 23:16
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
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
24 Mar 2019, 18:25
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. Answer: C
_________________
5star rated online GMAT quant self study course 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






