Author 
Message 
TAGS:

Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8123
GPA: 3.82

If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 01:37
Question Stats:
71% (02:09) correct 29% (02:35) wrong based on 95 sessions
HideShow timer Statistics
[ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 1 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Director
Joined: 20 Feb 2015
Posts: 738
Concentration: Strategy, General Management

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 02:02
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) we cannot have all 3 odd numbers because o+o+o =o therefore, one even prime number , which is 2 is always required . no of ways we can select 2 = 1 no of ways we can select the remaining = \(7c2\) = 21 no of ways we can select 3 = \(8c3\) = 56 p(sum is even) = \(\frac{21}{56}\)= \(\frac{3}{8}\) E



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4055
Location: Canada

If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 08:00
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) The first 8 prime numbers are: {2, 3, 5, 7, 11, 13, 17, 19} Notice that only one prime number is EVEN, and the remaining seven numbers are ODD. Also notice that the sum of 3 numbers will be EVEN only if one of the three selected numbers is 2 So, the question is really asking "What is the probability that 2 is among the three selected values?" There are many ways to answer this question. A superquick approach is to use the complement That is, P(Event A happening) = 1  P(Event A not happening) So, we can write: P(2 is among the three selected values) = 1  P(2 is NOT among the three selected values)P(2 is NOT among the three selected values)P(2 is NOT among the three selected values) = P(all three selected values are ODD) = P(1st number is odd AND 2nd number is odd AND 3rd number is odd) = P(1st number is odd) x P(2nd number is odd) x P(3rd number is odd) = 7/8 x 6/7 x 5/6 = 5/8So, P(2 is among the three selected values) = 1  5/8= 3/8 Answer: E Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Manager
Joined: 14 Jun 2018
Posts: 212

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 11:01
Since the sum is even , it must contain 2 as one of the numbers. Remaining numbers can be selected in 7c2 ways Probability = 7c2/8c3 = 3/8



Senior Manager
Joined: 04 Aug 2010
Posts: 492
Schools: Dartmouth College

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 11:01
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) First 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19. The sum will be even if 2 is among the three numbers selected, with the result that the sum = EVEN PRIME + ODD PRIME + ODD PRIME = EVEN. Since 3 NUMBERS are selected from 8 OPTIONS, the probability that 2 will be among the three numbers selected = \(\frac{3}{8}\).
_________________
GMAT and GRE Tutor Over 1800 followers GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and longdistance. For more information, please email me at GMATGuruNY@gmail.com.



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935

If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 12:09
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) \({\rm{first}}\,\,{\rm{8}}\,\,{\rm{primes}}\,\,\left\{ \matrix{ \,{\rm{first}} = 2 = {\rm{even}} \hfill \cr \,{\rm{7}}\,{\rm{others}}\,\, = \,\,{\rm{odd}}\,\,\,\,\,\left( {{\rm{it}}\,\,{\rm{does}}\,\,{\rm{not}}\,\,{\rm{matter}}\,{\rm{who}}\,\,{\rm{they}}\,\,{\rm{are}}!} \right) \hfill \cr} \right.\,\,\,\,\,\,\) \(? = {\mathop{\rm P}\nolimits} \left( {{\rm{3}}\,\,\underline {{\rm{different}}} \,\,{\rm{selected}}\,\,{\rm{have}}\,\,{\rm{sum}}\,\,{\rm{even}}} \right) = {\mathop{\rm P}\nolimits} \left( {{\rm{number}}\,{\rm{2}}\,\,{\rm{is}}\,\,{\rm{selected}},\,\,{\rm{other}}\,\,2\,\,{\rm{free}}} \right)\) \({\rm{total}} = C\left( {8,3} \right)\,\,\,{\rm{equiprobable}}\) \({\rm{favorable}}\,\,{\rm{ = }}\,\,{\rm{C}}\left( {7,2} \right)\,\,\,\,\left[ {2\,\,{\rm{had}}\,\,{\rm{already}}\,\,{\rm{stolen}}\,\,{\rm{a}}\,{\rm{place,}}\,\,{\rm{have}}\,\,{\rm{7}}\,\,{\rm{options}}\,\,{\rm{for}}\,\,{\rm{the}}\,{\rm{ other}}\,\,{\rm{two }}\,{\rm{numbers}}\,} \right]\) \(? = {{{{7!} \over {2!\,\,5!}}} \over {{{8!} \over {3!\,\,5!}}}} = {{7!\,\,3!} \over {2!\,\,8!}} = {3 \over 8}\) 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



VP
Joined: 07 Dec 2014
Posts: 1222

If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
02 Oct 2018, 13:58
[ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) total possibilities: 8*7*6=336 odd sum possibilities: 7*6*5=210 even sum possibilities: 336210=126 126/336=3/8 E



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8345
Location: United States (CA)

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
03 Oct 2018, 18:00
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(3\) different numbers are selected from the first \(8\) prime numbers, what is the probability that the sum of the \(3\) numbers selected is even? \(A. \frac{1}{4}\) \(B. \frac{1}{3}\) \(C. \frac{1}{2}\) \(D. \frac{2}{3}\) \(E. \frac{3}{8}\) We recall that odd + odd + odd = odd and that even + odd + odd = even. The first 8 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19. There are 8C3 = (8 x 7 x 6)/(3 x 2) = 56 ways to choose 3 of them. However, in order for the sum of the 3 numbers to be even, one of them must be 2 (the other two numbers can be any two of the remaining 7 odd primes). The number of ways of choosing 2 odd primes from 7 odd primes is 7C2 = (7 x 6)/2 = 21. Therefore, the probability that the sum of the 3 numbers selected is even is 21/56 = 3/8. Answer: E
_________________
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.



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8123
GPA: 3.82

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
04 Oct 2018, 00:57
=> Suppose \(p, q\) and \(r\) are prime numbers. In order for \(p + q + r\) to be even, one of them must equal \(2\), since \(2\) is the only even prime number. Once \(2\) has been selected, there are \(7\) prime numbers remaining from which to select \(2\) numbers. Thus, the number of selections in which the sum of the \(3\) numbers is even is 7C 2 = 21. The total number of selections is 8C 3 = 56. Thus, the probability that the sum of the three numbers is even is \(\frac{21}{56} = \frac{3}{8}.\) Therefore, the answer is E. Answer: E
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 1 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Manager
Joined: 17 Mar 2018
Posts: 70

Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
Show Tags
04 Oct 2018, 01:41
I went with complement method. Probability of all odd numbers being selected= 7/8*6/7*5/6=5/8 5/8th probability that sum of selected numbers is odd Probability of selecting the only even prime number= 15/8= 3/8 the probability that sum of selected number is even




Re: If 3 different numbers are selected from the first 8 prime numbers, wh
[#permalink]
04 Oct 2018, 01:41






