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

It is currently 17 Aug 2018, 00:24

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

Set S is the set of all prime integers between 0 and 20. If

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

Hide Tags

GMAT Club Legend
GMAT Club Legend
User avatar
B
Joined: 29 Jan 2005
Posts: 5092
Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 01 Sep 2006, 00:40
3
21
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

73% (02:38) correct 27% (02:49) wrong based on 467 sessions

HideShow timer Statistics

Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
(C) 17/28
(D) 3/4
(E) 301/336
Most Helpful Community Reply
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 778
Location: BULGARIA
  [#permalink]

Show Tags

New post 01 Sep 2006, 02:03
9
4
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28
General Discussion
Director
Director
User avatar
Joined: 28 Dec 2005
Posts: 727
  [#permalink]

Show Tags

New post 01 Sep 2006, 02:48
1
C for me.

P(product < 31) =1/8C3

P(sum is odd) = 7C3/8C3

Diff = 34/56 or 17/28
Senior Manager
Senior Manager
avatar
Joined: 14 Jul 2006
Posts: 280
Re: PS Sets  [#permalink]

Show Tags

New post 01 Sep 2006, 04:49
1
1
GMATT73 wrote:
Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
(C) 17/28
(D) 3/4
(E) 301/336


Ans C The only time the product will be less than 31 is when 2,3,5

1/(8C3)

The only time an even sum would occur is when 2 is included in the mix

so excluding 2 7C3/8C3

The difference will be C
GMAT Club Legend
GMAT Club Legend
User avatar
B
Joined: 29 Jan 2005
Posts: 5092
  [#permalink]

Show Tags

New post 01 Sep 2006, 04:50
1
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


Scintillating :cool Don't you just love it when we can reinforce multiple concepts in one problem?

1. Rule of primes
2. Adding odd and even integers
3. Triplets
4. Dependent probabilty
5. Combinatorics
6. Positive sum (absolute value)
Intern
Intern
avatar
Joined: 14 Feb 2012
Posts: 25
Re:  [#permalink]

Show Tags

New post 31 Oct 2013, 02:08
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


Im not understanding how you come to 35/56?

Can someone please explain?
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1851
Concentration: Finance
GMAT ToolKit User
Re:  [#permalink]

Show Tags

New post 25 Nov 2013, 12:55
1
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


You got the answer but your procedure is flawed. You must INCLUDE 2 not exclude since you want the sum to be odd. Anyways you get the same combinatorics fraction.

Cheers
J :)
Retired Moderator
avatar
B
Joined: 17 Sep 2013
Posts: 369
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
GMAT ToolKit User Premium Member Reviews Badge
Re: Re:  [#permalink]

Show Tags

New post 07 May 2014, 02:34
jlgdr wrote:
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


You got the answer but your procedure is flawed. You must INCLUDE 2 not exclude since you want the sum to be odd. Anyways you get the same combinatorics fraction.

Cheers
J :)


2+3+5= 10...?
3 numbers here..so all 3 odd is necessary

FYI if we use 2 here..i.e 2 is a necessary filler..then the no of ways you can get an even sum is 21..and probability is 7C2/8C3=21/56
Correct me if I am wrong
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..:P

Intern
Intern
avatar
Joined: 28 Jun 2015
Posts: 6
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 23 Aug 2015, 23:48
2
1
S = 2,3,5,7,11,13,17,19
Size of S = 8

first number can be chosen in 8 ways
2nd number in 7 ways
3rd number 6 ways

so total outcomes = 8 x 7 x 6 = 336 possible triplets

for the sum to be odd all three must be odd = 7 x 6 x 5 = 210 ways

prob of odd = 210/336 = 5/8

prod less than 31 can only be possible if we have 2,3,5 this can be done in 3x2x1 ways = 6 ways
prob (sum<31) = 6/8x7x6 = 1/56

diff = 5/8 - 1/56 = 34/56 = 17/28
Manager
Manager
avatar
Joined: 07 Apr 2015
Posts: 176
Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 24 Aug 2015, 01:47
1
1
I did it this way:

Primes between 0 and 20:
2, 3, 5, 7, 11, 13, 17, 19

I: Probability that the product of (x,yz) < 31:
Only possible way is 2*3*5 = 30, all other products are >31

=> \(1/8 * 1/7 * 1/6 * 3! = 3*2*1/336 = 1/56\)

II: Probability that the sum of (x+y+z) = odd:
There are numerous possibilities as long as 2 (the only even number) is not included.

=> \(7/8 * 6/7 * 5/6 = 5/8\)

III: I - II:

=> \(1/56 - 5/8 = |-17/28|\)
Intern
Intern
avatar
B
Joined: 27 Apr 2015
Posts: 33
Location: India
GMAT 1: 730 Q50 V40
WE: Operations (Telecommunications)
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 31 Aug 2015, 22:23
1
waltiebikkiebal wrote:
BG wrote:
The primes between 1-20 are 2,3,5,7,11,13,17,19 or 8 in total then 8C3=56 or we can form 56 triplets. The product of 2,3,5 only is less than 31 so the prob that the product is less than 31 is 1/56 . The prob that the sum is odd is 7C3/8C3 . Exclude 2 , because it is even and will make the sum even and select 3 out of 7( Odd only) the prob is 35/56. The required prob is 35/56-1/56=34/56=17/28


Im not understanding how you come to 35/56?

Can someone please explain?



We have a total of 8 prime nos between 1 and 20. To have an odd sum, the three nos should be odd. Out of the total 8 prime nos, there are 7 odd nos, so the no of ways to select 3 nos out of 7 would be 7C3.
Hence, the probability would be 7C3/8C3 = 35/56
Manager
Manager
avatar
G
Joined: 14 Oct 2012
Posts: 175
Premium Member Reviews Badge
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 27 Aug 2016, 17:01
1
Hello Friends,
I am really confused by this question and here's why: it states "the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same :?:



Thanks for your help in advance.
Intern
Intern
User avatar
Status: Pursuit of Happiness
Joined: 10 Sep 2016
Posts: 30
Location: United States (IL)
Concentration: Finance, Economics
Schools: HBS '19, CBS '19
GMAT 1: 590 Q44 V27
GMAT 2: 690 Q50 V34
GPA: 3.94
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 16 Oct 2016, 16:20
3
manishtank1988 wrote:
Hello Friends,
I am really confused by this question and here's why: it states "the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same :?:



Thanks for your help in advance.


Hi there,

Don't worry about "the product less than 31" part. Just focus on what the question asks: The probability that the sum will be odd.
In order for the sum to be odd, you need either 2 evens and one odd, or 3 odds. Since all the prime numbers only have one even number that is 2, 2 evens and one odd is not an option. So just focus on 3 odds.
There are only 7 odds prime number between 0 and 20. Hence, the answer is 7C3/8C3.
_________________

If you find this post hepful, please press +1 Kudos

Intern
Intern
User avatar
Status: Pursuit of Happiness
Joined: 10 Sep 2016
Posts: 30
Location: United States (IL)
Concentration: Finance, Economics
Schools: HBS '19, CBS '19
GMAT 1: 590 Q44 V27
GMAT 2: 690 Q50 V34
GPA: 3.94
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 16 Oct 2016, 20:35
2
manishtank1988 wrote:
Hello Friends,
I am really confused by this question and here's why: it states "the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd"

So shouldn't this mean that the triplet should satisfy BOTH conditions?
I thought like this and did following:
(i) For product of a*b*c < 31 we need 2 BUT
(ii) for sum to be odd ie a+b+c = odd we need 3 odd numbers.
Thus, P(a*b*c < 31) = 0 and P(a+b+c = o) = (7C3)/(8C3) = 35/56 = 5/8.
Clearly, this doesn't seem to be correct according to answer choices. So my question to you'll is:

q) how did you arrive at the conclusion that triplets for both conditions shouldn't be the same :?:



Thanks for your help in advance.


Hi again,

I apologize -- I just noticed that part of what I had written is missing.

So I was saying that the probability that the sum of the 3 numbers is odd is 7C3/8C3.
Now let's focus on the remaining of the question: the positive difference between the above probability - 7C3/8C3 - and the probability that the product of these 3 numbers is a number less than 31.
There is only one set of numbers from the list that gives a product less than 31. That set of number is 2,3,5 - 2*3*5=30. Any other set would give a larger number. That being said, substract 3C3/8C3 from 7C3/8C3 and you get the answer.

Hope it helps.

--
Still need 19 Kudos
_________________

If you find this post hepful, please press +1 Kudos

Director
Director
User avatar
G
Joined: 26 Oct 2016
Posts: 656
Location: United States
Concentration: Marketing, International Business
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 05 Jan 2017, 21:01
If set S is the set of all prime integers between 0 and 20 then:
S = {2, 3, 5, 7, 11, 13, 17, 19}
Let’s start by finding the probability that the product of the three numbers chosenis a number less than 31. To keep the product less than 31, the three numbers must be 2, 3 and 5. So, what is the probability that the three numbers chosen will be some combination of 2, 3, and 5?
Here’s the list all possible combinations of 2, 3, and 5:
case A: 2, 3, 5
case B: 2, 5, 3
case C: 3, 2, 5
case D: 3, 5, 2
case E: 5, 2, 3
case F: 5, 3, 2
This makes it easy to see that when 2 is chosen first, there are two possible combinations. The same is true when 3 and 5 are chosen first. The probability of drawing a 2, AND a 3, AND a 5 in case A is calculated as follows (remember, when calculating probabilities, AND means multiply):
case A: (1/8) x (1/7) x (1/6) = 1/336
The same holds for the rest of the cases.
case B: (1/8) x (1/7) x (1/6) = 1/336
case C: (1/8) x (1/7) x (1/6) = 1/336
case D: (1/8) x (1/7) x (1/6) = 1/336
case E: (1/8) x (1/7) x (1/6) = 1/336
case F: (1/8) x (1/7) x (1/6) = 1/336
So, a 2, 3, and 5 could be chosen according to case A, OR case B, OR, case C, etc. The total probability of getting a 2, 3, and 5, in any order, can be calculated as follows (remember, when calculating probabilities, OR means add):
(1/336) + (1/336) + (1/336) + (1/336) + (1/336) + (1/336) = 6/336
Now, let’s calculate the probability that the sum of the three numbers is odd. In order to get an odd sum in this case, 2 must NOT be one of the numbers chosen. Using the rules of odds and evens, we can see that having a 2 would give the following scenario:
even + odd + odd = even
So, what is the probability that the three numbers chosen are all odd? We would need an odd AND another odd, AND another odd:
(7/8) x (6/7) x (5/6) = 210/336
The positive difference between the two probabilities is:
(210/336) – (6/336) = (204/336) = 17/28
The correct answer is C.
_________________

Thanks & Regards,
Anaira Mitch

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 7733
Premium Member
Re: Set S is the set of all prime integers between 0 and 20. If  [#permalink]

Show Tags

New post 01 Aug 2018, 08:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: Set S is the set of all prime integers between 0 and 20. If &nbs [#permalink] 01 Aug 2018, 08:35
Display posts from previous: Sort by

Set S is the set of all prime integers between 0 and 20. If

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

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.