NEW HERE? LEARN MORE
closeclose
Last visit was: 24 Apr 2024, 16:46 It is currently 24 Apr 2024, 16:46
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Three distinct integers are selected at random between 1 and 2016, inc

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618811 [3]
Given Kudos: 81588
Send PM
CAT Tests Forum Quiz Mode
Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   25 Apr 2019, 03:50
1
Kudos
2
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

60% (02:22) correct 40% (02:06) wrong based on 90 sessions
Hide Show timer Statistics
Three distinct integers are selected at random between 1 and 2016, inclusive (without replacement). Which of the following is a correct statement about the probability p that the product of the three integers is odd?

(A) p < 1/8
(B) p = 1/8
(C) 1/8 < p < 1/3
(D) p = 1/3
(E) p > 1/3
ShowHide Answer
Official Answer
A
avatar
S
aliakberza
Intern
Intern
Joined: 11 Feb 2018
Posts: 45
Own Kudos [?]: 105 [2]
Given Kudos: 148
Send PM
Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   25 Jun 2019, 11:31
2
Kudos
greenisthecolor wrote:
I took way too much time to solve this. Could someone tell me a quick way to solve this?


I tried to solve and reduce the fractions too, but then realized that should there have been replacement (i.e. numbers would be brought back to 2016 after each selection) the probability would have been exactly 1/2 * 1/2 * 1/2 = 1/8.

Now as a general rule of thumb when the same number is added or subtracted to/from both the numerator and denominator, the resulting fraction will always have a greater or lower value respectively. Because there is no replacement each time a number is selected, we know that the absolute value of the fraction is decreasing after each selection. If with replacement we know the probability would have been 1/8, then without replacement it should be \(p<\frac{1}{8}\).

Hope this helps. :)
User avatar
L
IanStewart
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9242 [2]
Given Kudos: 91
 Q51  V47
Send PM
Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   25 Jun 2019, 11:45
2
Kudos
Expert Reply
aliakberza wrote:
Now as a general rule of thumb when the same number is added or subtracted to/from both the numerator and denominator, the resulting fraction will always have a greater or lower value respectively.


That is true only when the overall value of the fraction is less than 1 (and when everything is positive). If you instead start with a fraction like 3/2, and you add 1 to the numerator and denominator, you'll see that the value drops. And if any numbers might be negative, the rules are a lot more complicated.
avatar
B
TornDilemma
Intern
Intern
Joined: 24 Apr 2019
Posts: 12
Own Kudos [?]: 92 [3]
Given Kudos: 2
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   25 Apr 2019, 04:11
1
Kudos
2
Bookmarks
Bunuel wrote:
Three distinct integers are selected at random between 1 and 2016, inclusive (without replacement). Which of the following is a correct statement about the probability p that the product of the three integers is odd?

(A) p < 1/8
(B) p = 1/8
(C) 1/8 < p < 1/3
(D) p = 1/3
(E) p > 1/3


No of Odd Integers between 1 and 2016=1008

Probability that thier product is odd=\(\frac{1008C3}{2016C3}\)

\(\frac{1008*1007*1006}{2016*2015*2014}=\frac{1006}{2*2*2015}\).

Now, \(\frac{1}{2015}<\frac{1}{2012}\)
So,\(\frac{1006}{2*2*2015}<\frac{1006}{2*2*2012}\) or \(p<\frac{1}{8}\).

IMO, Option A.
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18756
Own Kudos [?]: 22047 [1]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   27 Jun 2019, 18:45
1
Kudos
Expert Reply
Bunuel wrote:
Three distinct integers are selected at random between 1 and 2016, inclusive (without replacement). Which of the following is a correct statement about the probability p that the product of the three integers is odd?

(A) p < 1/8
(B) p = 1/8
(C) 1/8 < p < 1/3
(D) p = 1/3
(E) p > 1/3


In order for the product of the three integer to be odd, each integer must be odd. There are 1008 odd integers from 1 to 2016, inclusive, so P(odd and odd and odd) = 1008/2016 x 1007/2015 x 1006/2014.

We see that 1008/2016 is exactly 1/2. Although 1007/2015 and 1006/2014 are not exactly 1/2, they are very close to 1/2. Therefore, the probability of selecting 3 odd numbers is approximately ½ x ½ x ½ = ⅛. Finally, because 1007/2015 and 1006/2014 are each slightly less than ½, the probability sought will be slightly less than ⅛.

Answer: A
User avatar
B
greenisthecolor
Intern
Intern
Joined: 03 Apr 2018
Posts: 27
Own Kudos [?]: 16 [0]
Given Kudos: 45
Location: India
Schools: Tepper '23 (WL) Melbourne (A$) ISB'22 (A) McCombs '23 (A)
GMAT 1: 720 Q49 V39
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   24 Jun 2019, 10:55
I took way too much time to solve this. Could someone tell me a quick way to solve this?
User avatar
L
nick1816
Retired Moderator
Joined: 19 Oct 2018
Posts: 1878
Own Kudos [?]: 6296 [0]
Given Kudos: 704
Location: India
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   24 Jun 2019, 12:18
Product of integers is odd, when all of them are odd numbers.
There are 1008 odd numbers from 1 to 2016

the probability p that the product of the three integers is odd= (\(\frac{1008}{2016}\))*(\(\frac{1007}{2015}\))*(\(\frac{1006}{2014}\))
p=\(\frac{(1008*1007*1006)}{(2016*2015*2014)}\)
p={\(\frac{1}{4}\)}*(\(\frac{1006}{2015}\))

\(\frac{1006}{2015}\)<\(\frac{1007.5}{2015}\)
Hence p= \(\frac{1}{4*2.xyz}\)
p<1/8





greenisthecolor wrote:
I took way too much time to solve this. Could someone tell me a quick way to solve this?
User avatar
L
IanStewart
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9242 [0]
Given Kudos: 91
 Q51  V47
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   24 Jun 2019, 13:24
Expert Reply
The probability our first selection is odd is 1/2. But if we're doing things without replacement, once we remove that first odd number, the probability the next is odd will be slightly less than 1/2, because now less than half of our numbers are odd. And the probability the third is odd again is even smaller than that. So the probability all three are odd will be less than (1/2)(1/2)(1/2) = 1/8 and A is the answer.
avatar
S
aliakberza
Intern
Intern
Joined: 11 Feb 2018
Posts: 45
Own Kudos [?]: 105 [0]
Given Kudos: 148
Send PM
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink] New post   25 Jun 2019, 11:48
IanStewart wrote:
aliakberza wrote:
Now as a general rule of thumb when the same number is added or subtracted to/from both the numerator and denominator, the resulting fraction will always have a greater or lower value respectively.


That is true only when the overall value of the fraction is less than 1 (and when everything is positive). If you instead start with a fraction like 3/2, and you add 1 to the numerator and denominator, you'll see that the value drops. And if any numbers might be negative, the rules are a lot more complicated.


Correct. Thanks for the input. IanStewart :)
GMAT Club Bot
gmatclubot
Re: Three distinct integers are selected at random between 1 and 2016, inc [#permalink]
25 Jun 2019, 11:48
Moderators:
Bunuel
Math Expert
92900 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions