GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 22:14 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.  If three prime number are randomly selected from prime numbers less

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

Hide Tags

Senior Manager  V
Joined: 22 Feb 2018
Posts: 419
If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

6
14 00:00

Difficulty:   95% (hard)

Question Stats: 54% (02:32) correct 46% (02:33) wrong based on 334 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%

_________________
Good, good Let the kudos flow through you
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3565
If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

7
1
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 sum

There 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

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.

Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver
Intern  B
Joined: 01 Jan 2018
Posts: 43
Re: If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

5
1
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.
General Discussion
Intern  B
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

1
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  P
Joined: 07 Dec 2014
Posts: 1224
Re: If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

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=720-504=216
216/720=.3=30%
C
Intern  B
Joined: 02 Dec 2017
Posts: 1
Re: If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

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  S
Joined: 10 Jun 2014
Posts: 83
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

1
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  D
Joined: 09 Mar 2016
Posts: 1230
If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

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 $$9C3 = \frac{9*8*7}{3!(9-3)} = 84$$

Two odd numbers to choose from 9 numbers: $$9C2 = \frac{9*8}{2!(9-2)} = 36$$

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

$$7C1 = \frac{7!}{(7-1)}= 7$$

$$Two Odd + One Even = 9C2 +7C1 = 36+7 =43$$

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 V
Joined: 26 Feb 2016
Posts: 3335
Location: India
GPA: 3.12
If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

1
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 $$9C3 = \frac{9*8*7}{3!(9-3)} = 84$$

Two odd numbers to choose from 9 numbers: $$9C2 = \frac{9*8}{2!(9-2)} = 36$$

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

$$7C1 = \frac{7!}{(7-1)}= 7$$

$$Two Odd + One Even = 9C2 +7C1 = 36+7 =43$$

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 dave13

You 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  V
Joined: 22 Feb 2018
Posts: 419
Re: If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

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

judsonvpires
C stands for combinations

$$nCk = \frac{n!}{(n-k)!*k!}$$

$$10C3 = \frac{10!}{(10-3)!*3!} =\frac{10!}{7!*3!}$$
_________________
Good, good Let the kudos flow through you
Intern  B
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

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 $$9C3 = \frac{9*8*7}{3!(9-3)} = 84$$

Two odd numbers to choose from 9 numbers: $$9C2 = \frac{9*8}{2!(9-2)} = 36$$

Now we are left with 7 numbers so there is only one even number hence I can choose only one number from 7 ---- >

$$7C1 = \frac{7!}{(7-1)}= 7$$

$$Two Odd + One Even = 9C2 +7C1 = 36+7 =43$$

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 G
Status: Private GMAT Tutor
Joined: 22 Oct 2012
Posts: 142
Location: India
Concentration: Economics, Finance
Schools: IIMA (A)
GMAT 1: 780 Q51 V47 If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

Top Contributor
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 G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815
Re: If three prime number are randomly selected from prime numbers less  [#permalink]

Show Tags

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%.

_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@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: If three prime number are randomly selected from prime numbers less   [#permalink] 03 Sep 2018, 18:59
Display posts from previous: Sort by

If three prime number are randomly selected from prime numbers less

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  