GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 25Crafting an Impactful Resume: Balancing Detail with Brevity
10:00 AM EDT
-10:30 AM EDT
Showcase your achievements through impactful storytelling! Get ready for an LIVE session on April 25, Learn how to craft compelling narratives on your resume that resonate with admission committees.
Apr 26Get a Top GRE Score with Target Test Prep
12:00 PM EDT
-01:00 PM EDT
The Target Test Prep course represents a quantum leap forward in GRE preparation, a radical reinterpretation of the way that students should study. In fact, more and more Target Test Prep students are scoring 329 and above on the GRE.
The only contents of a container are 10 disks that are each
[#permalink]
Updated on: 01 Feb 2019, 01:16
Updated on: 01 Feb 2019, 01:16
1
1
11
Bookmarks
00:00
Question Stats:
60% (02:15) correct
40% (02:33) wrong based on 25 sessions
Hide Show timer Statistics
The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?
A) \(\frac{1}{7}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{2}\)
E) \(\frac{15}{28}\)
A) \(\frac{1}{7}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{2}\)
E) \(\frac{15}{28}\)
Originally posted by CAMANISHPARMAR on 21 Jan 2019, 03:08.
Last edited by chetan2u on 01 Feb 2019, 01:16, edited 1 time in total.
Last edited by chetan2u on 01 Feb 2019, 01:16, edited 1 time in total.
Corrected the OA
Re: The only contents of a container are 10 disks that are each
[#permalink]
21 Jan 2019, 21:35
21 Jan 2019, 21:35
5
This should be a select one multiple choice question.
There are altogether 10 integers namely disk labelled 1 to 10 inclusive therefore total number of outcomes for choosing 4 integers out of 10 is \(10C4 = \frac{10*9*8*7}{4*3*2} = 210\)
Now lets look at the ways in which we can chose 4 integers such that the range becomes 7.
There are only 3 possibilities how a range of 7 could occur when 4 numbers are chosen
1st. \((1,8)\): If among the 4 number chosen 2 of the numbers are 1 and 8 and the rest of the 2 numbers are from {2,3,4,5,6,7}. Notice that 9 cannot be chosen because it results in a range of 8.
2nd. \((2,9)\): Rest of the 2 numbers should be from the set {3,4,5,6,7,8}
3rd \((3,10)\): Rest of the 2 numbers should be from the set {4,5,6,7,8,9}
Now let us take the case for any 1 of 3 possible option.
(2,9)
There are 4 slots to chose the number. First 2 should be (2,9)
Hence 2c2 = 1
Next 2 numbers should be from the set of remaining 6 integers {3,4,5,6,7,8}
Hence \(6c2 = 15\)
Therefore for 1 of the possibility the required outcome = 15 ways
There are 3 possibilities therefore total possible ways = \(15*3 = 45\)
Therefore the probability is \(\frac{45}{210}\) which reduces to \(\frac{3}{14}\)
There are altogether 10 integers namely disk labelled 1 to 10 inclusive therefore total number of outcomes for choosing 4 integers out of 10 is \(10C4 = \frac{10*9*8*7}{4*3*2} = 210\)
Now lets look at the ways in which we can chose 4 integers such that the range becomes 7.
There are only 3 possibilities how a range of 7 could occur when 4 numbers are chosen
1st. \((1,8)\): If among the 4 number chosen 2 of the numbers are 1 and 8 and the rest of the 2 numbers are from {2,3,4,5,6,7}. Notice that 9 cannot be chosen because it results in a range of 8.
2nd. \((2,9)\): Rest of the 2 numbers should be from the set {3,4,5,6,7,8}
3rd \((3,10)\): Rest of the 2 numbers should be from the set {4,5,6,7,8,9}
Now let us take the case for any 1 of 3 possible option.
(2,9)
There are 4 slots to chose the number. First 2 should be (2,9)
Hence 2c2 = 1
Next 2 numbers should be from the set of remaining 6 integers {3,4,5,6,7,8}
Hence \(6c2 = 15\)
Therefore for 1 of the possibility the required outcome = 15 ways
There are 3 possibilities therefore total possible ways = \(15*3 = 45\)
Therefore the probability is \(\frac{45}{210}\) which reduces to \(\frac{3}{14}\)
General Discussion
Re: The only contents of a container are 10 disks that are each
[#permalink]
31 Jan 2019, 23:40
31 Jan 2019, 23:40
2
Expert Reply
In case pure combinatorics is too tough, this is a bit of a longer way, but it explains the how and why
there are only 3 possibilities where the range can be 7.
(1,8) (2,9) (3,10)
If we choose (1,8),
the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9
now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168
But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.
12*1/168= 1/14
If we choose (2,9)
the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9
now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
If we choose (3,10)
the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.
now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B
1/90+1/120+1/168= 8/315
Any pointers on how to bridge this gap?
there are only 3 possibilities where the range can be 7.
(1,8) (2,9) (3,10)
If we choose (1,8),
the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9
now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168
But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.
12*1/168= 1/14
If we choose (2,9)
the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9
now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
If we choose (3,10)
the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.
now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B
1/90+1/120+1/168= 8/315
Any pointers on how to bridge this gap?
