Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 22 Feb 2018
Posts: 415

If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
15 Jun 2018, 04:14
Question Stats:
55% (02:33) correct 45% (02:33) wrong based on 335 sessions
HideShow timer Statistics
If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even? A) 10% B) 27% C) 30% D) 36.5% E) 42%
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Good, good Let the kudos flow through you




Senior SC Moderator
Joined: 22 May 2016
Posts: 3655

If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
15 Jun 2018, 08:10
Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% Find the probability of "NOT odd," a task that is much easier than that in the prompt, and subtract from 1. There are 10 prime numbers that are less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 9 of 10 are odd numbers The sum of any three numbers NOT including 2 will be odd.* The sum of any 3 odds will be odd: (O + O) = E (first two odd #s), and then (E + O) (plus third odd #) = Odd Example: (3 + 5) = 8, and then (8 + 7) = 15 Use the probability complement rule Even = not Odd Odd = not Even P(E) + P(not E) = 1 P(Even Sum) + P(Odd Sum, which is Not E) = 1 P(Even) = 1  P(Odd) Probability of odd sumThere are 10 numbers to pick from. 9 are odd. P of First Pick odd: \(\frac{9}{10}\) Only 9 numbers remain. 8 are odd. P of Second Pick odd: \(\frac{8}{9}\) 8 numbers remain. 7 are odd P of Third Pick odd: \(\frac{7}{8}\) P(Odd Sum): \((\frac{9}{10}*\frac{8}{9}*\frac{7}{8})=\frac{7}{10}=0.70\) P(Even Sum) = \((1  0.70) = 0.30\) = 30 percent Answer C If need be, try adding numbers to find the pattern: (2+3+5) = 10 ... EVEN sum (3+5+7) = 15 ... ODD sum (5+7+11) = 23 ... ODD sum (7+11+13) = 31 ... ODD sum Pattern: 3 odds sum to ODD. Leave 2 out of the picture until the end.
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has  Margaret Mead




Intern
Joined: 01 Jan 2018
Posts: 43

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
15 Jun 2018, 08:13
Only one prime number is even and all other primes are odd. So for the sum to be even : two primes should be odd and the other one should be even prime.
No. of prime number (2,3,5,7,11,13,17,19, 23, 29) = 10 numbers.
Even prime can be selected in 1C1 ways = 1 Odd prime can be selected in 9C2 ways = 36
Total ways in which 3 numbers can be selected = 10C3 = 120
Therefore percentage of sum of 3 primes being even = (1*36)/120 = (3/10)*100 = 30%
OA  C.




Intern
Status: No Progress without Struggle
Joined: 04 Aug 2017
Posts: 42
Location: Armenia
GPA: 3.4

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
24 Aug 2018, 09:05
To have the sum as an even number one of the numbers should be even. ODD+ODD+EVEN=EVEN ODD+EVEN+ODD=EVEN Logically we can conclude that the probability that the prime number 2 will be selected out of 9 other odd numbers is relatively low.
_________________
Seryozha Sargsyan 21
Contact: sargsyanseryozha@gmail.com
What you think, you become, What you feel, you attract, What you imagine, you create.



VP
Joined: 07 Dec 2014
Posts: 1222

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
24 Aug 2018, 13:05
Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% total possibilities=10*9*8=720 total odd possibilities=9*8*7=504 total even possibilities=720504=216 216/720=.3=30% C



Intern
Joined: 02 Dec 2017
Posts: 1

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
24 Aug 2018, 13:49
pr3mk5: Your answer is brilliantly short and intuitive  thanks. It contained the following: "Even prime can be selected in 1C1 ways = 1 Odd prime can be selected in 9C2 ways = 36 Total ways in which 3 numbers can be selected = 10C3 = 120" Could you please elaborate more on how to get the "total ways"  > didnt get the "10C3" (actually dont know what's the "C" in there) Thanks



Manager
Joined: 10 Jun 2014
Posts: 82
Location: India
Concentration: Operations, Finance
WE: Manufacturing and Production (Energy and Utilities)

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
29 Aug 2018, 12:14
There are 10 prime numbers that are less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Only 2 is even , rest are odd Now if the sum of three prime numbers is even , 2 must be selected in one of the three number otherwise the result will always be ODD as ODD +ODD+ODD = ODD Now there are three ways  1) 2 AND any odd number AND any odd number OR 2) any odd number AND 2 AND any odd number OR 3) any odd number AND any odd number AND 2 Probability  (1/10)*(9/9)*(8/8) + (9/10)*(1/9)*(8/8) + (9/10)*(8/9)*(1/8) = 3/10=0.3 Ans C
_________________
KUDOS PLEASE IF IT HELPED



VP
Joined: 09 Mar 2016
Posts: 1229

If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
30 Aug 2018, 10:17
Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% 29, 23, 19, 17, 13, 11, 7, 5, 2 Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(93)} = 84\) Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(92)} = 36\) Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7  > \(7C1 = \frac{7!}{(71)}= 7\) \(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\) So probability is \(\frac{43}{84}\) = ca. \(0.5\) hello there pushpitkc  what, where and why ? it looks like a logicaly "correct" solution but i got incorrect answer what did i do wrong



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3308
Location: India
GPA: 3.12

If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
30 Aug 2018, 11:39
dave13 wrote: Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% 29, 23, 19, 17, 13, 11, 7, 5, 2 Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(93)} = 84\) Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(92)} = 36\) Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7  > \(7C1 = \frac{7!}{(71)}= 7\) \(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\) So probability is \(\frac{43}{84}\) = ca. \(0.5\) hello there pushpitkc  what, where and why ? it looks like a logicaly "correct" solution but i got incorrect answer what did i do wrong Hey dave13You did everything right except for this  there are 10 prime numbers less than 30. The total possibilities will change accordingly. This is what the correct solution would look like Listing all primes less than 30  29,23,19,17,13,11,7,5,3,2 Total ways of selecting 3 numbers from 10:\(C_3^{10} = \frac{10*9*8}{3*2*1} = 120\) There is only one possibility to choose 3 prime numbers and have an even sum, when we have 2 odd and 1 even prime number as the 3 primes need to be different Total possibility of choosing 2 odd primes from 9: \(C_3^{9} = \frac{9*8}{2*1} = 36\) There is only one possibility of choosing an even prime number. Therefore, Probability is \(\frac{36}{120} = 30\)% (Option C)Hope this clears your confusion!
_________________
You've got what it takes, but it will take everything you've got



Senior Manager
Joined: 22 Feb 2018
Posts: 415

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
01 Sep 2018, 07:02
judsonvpires wrote: pr3mk5: Your answer is brilliantly short and intuitive  thanks. It contained the following: "Even prime can be selected in 1C1 ways = 1 Odd prime can be selected in 9C2 ways = 36 Total ways in which 3 numbers can be selected = 10C3 = 120" Could you please elaborate more on how to get the "total ways"  > didnt get the "10C3" (actually dont know what's the "C" in there) Thanks judsonvpiresC stands for combinations \(nCk = \frac{n!}{(nk)!*k!}\) \(10C3 = \frac{10!}{(103)!*3!} =\frac{10!}{7!*3!}\)
_________________
Good, good Let the kudos flow through you



Intern
Status: No Progress without Struggle
Joined: 04 Aug 2017
Posts: 42
Location: Armenia
GPA: 3.4

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
02 Sep 2018, 06:43
dave13 wrote: Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% 29, 23, 19, 17, 13, 11, 7, 5, 2 Total number of ways to choose any three numbers from nine \(9C3 = \frac{9*8*7}{3!(93)} = 84\) Two odd numbers to choose from 9 numbers: \(9C2 = \frac{9*8}{2!(92)} = 36\) Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7  > \(7C1 = \frac{7!}{(71)}= 7\) \(Two Odd + One Even = 9C2 +7C1 = 36+7 =43\) So probability is \(\frac{43}{84}\) = ca. \(0.5\) hello there pushpitkc  what, where and why ? it looks like a logicaly "correct" solution but i got incorrect answer what did i do wrong The simple logic here is to understand that in reality there are two cases here. As we have only one even number in the list of prime numbers (only 2), there can be only two cases to choose three prime numbers: 1st case: odd+odd+odd 2nd case: odd+odd+even So, to make our calculations easier, we can subtract the 1st case from the total or 1 to get the 2nd case. In order to calculate the possibilities of the 1st case, we should do a simple probability calculation. As there are ten prime numbers in that range: 2,3,5,7,11,13,17,19,23,29. 9/10*8/9*7/8=504/720 And finally subtract 504/720 from the total, 1 to get 0.3, which in percent form is 30%.
_________________
Seryozha Sargsyan 21
Contact: sargsyanseryozha@gmail.com
What you think, you become, What you feel, you attract, What you imagine, you create.



GMAT Tutor
Status: Private GMAT Tutor
Joined: 22 Oct 2012
Posts: 144
Location: India
Concentration: Economics, Finance

If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
02 Sep 2018, 09:36
Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% There are 10 prime numbers below 30 and only one such number is even. For the sum of three selected numbers to be even, we have to select the even prime number (i.e. 2). If we don't select it, the sum will be odd. And if we select it, doesn't matter which other numbers we choose, the sum will be even. So, essentially, the probability asked in the question is the same as the probability that we'll select 2 as one of the three numbers. So, we have to select 2 as one of the numbers and the other two numbers can be any of the other 9 prime numbers. So, total ways to select such a set of three numbers out of a set of 10 prime numbers is 1C1 * 9C2. Total number of ways to select 3 numbers out of 10 numbers is 10C3. The required probability is \(\frac{9C2}{10C3}\) i.e. 0.3. I hope it helps.  CJ
_________________



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2812

Re: If three prime number are randomly selected from prime numbers less
[#permalink]
Show Tags
03 Sep 2018, 18:59
Princ wrote: If three prime numbers are randomly selected from the prime numbers less than 30 and no prime number can be chosen more than once, What is the probability that sum of the three prime numbers selected will be even?
A) 10% B) 27% C) 30% D) 36.5% E) 42% The prime numbers less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 The number of ways of picking any 3 of them is 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2) = 120. Our goal is to add 3 primes to obtain an even sum. Note that if all 3 prime numbers are odd, we will never obtain an even sum, so we know that one of the 3 primes chosen must be the number 2. So, from the remaining 9 primes, we need to pick any two of them. The number of ways of picking 2 primes from 9 is 9C2 = 9!/(2! x 7!) = (9 x 8)/2 = 36. Therefore, the probability is 36/120 = 3/10 = 30%. 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: If three prime number are randomly selected from prime numbers less
[#permalink]
03 Sep 2018, 18:59






