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.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.
In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
Updated on: 30 Jan 2019, 11:46
5
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
37% (02:24) correct 63% (02:10) wrong based on 59 sessions
HideShow timer Statistics
In a school of 900 students, all of them are coded with a distinct three-digit number from 100 to 999. If two students are select at random, what is the probability that exactly two digits of their codes are same?
A. \(^{81}C_2/^{900}C_2\) B. \(^{90}C_2/^{900}C_2\) C. \(^{162}C_2/^{900}C_2\) D. \(^{243}C_2/^{900}C_2\) E. \(^{270}C_2/^{900}C_2\)
Re: In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
30 Jan 2019, 05:01
EgmatQuantExpert wrote:
In a class of 100 students, all of them are coded with a distinct three-digit number from 100 to 199. If two students are select at random, what is the probability that exactly two digits of their codes are same?
A. \(^{81}C_2/^{100}C_2\) B. \(^{90}C_2/^{100}C_2\) C. \(^{162}C_2/^{100}C_2\) D. \(^{243}C_2/^{100}C_2\) E. \(^{270}C_2/^{100}C_2\)
You will have to look into the question again. How can the probability be greater than 1. Choices C to E are all above 1. Probability >1..Impossible
_________________
In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
Updated on: 31 Jan 2019, 02:06
total students = 900
total 27 such no would be there which have two digits exactly same in each series for eg 100 to 200 100,101,110,112to119,121,122,131,133,141,144,151,155,161,166,171,177,181,188,191,199 so in total we have 9 such series ; 100-200,200-300,300-400,400-500,500-600,600-700,700-800,800-900,900-1000
27*9 = 243 P ()= 243c2/900c2 IMO D
EgmatQuantExpert wrote:
In a class of 100 students, all of them are coded with a distinct three-digit number from 100 to 199. If two students are select at random, what is the probability that exactly two digits of their codes are same?
A. \(^{81}C_2/^{100}C_2\) B. \(^{90}C_2/^{100}C_2\) C. \(^{162}C_2/^{100}C_2\) D. \(^{243}C_2/^{100}C_2\) E. \(^{270}C_2/^{100}C_2\)
Originally posted by Archit3110 on 30 Jan 2019, 06:58.
Last edited by Archit3110 on 31 Jan 2019, 02:06, edited 1 time in total.
Re: In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
30 Jan 2019, 20:46
EgmatQuantExpert wrote:
In a school of 900 students, all of them are coded with a distinct three-digit number from 100 to 999. If two students are select at random, what is the probability that exactly two digits of their codes are same?
A. \(^{81}C_2/^{900}C_2\) B. \(^{90}C_2/^{900}C_2\) C. \(^{162}C_2/^{900}C_2\) D. \(^{243}C_2/^{900}C_2\) E. \(^{270}C_2/^{900}C_2\)
I'm guessing E. Here's my reasoning:
We have a total of 270 numbers that can have matching two digits.
100 -> 200 -> 300 -> 400....-> 900. We have a total of 10 ways we can match each number (00, 11, 22, 33... 99). Our numbers can range from 100 -> 999. This means we have 9 hundreds values we can use. We also have 3 columns that can provide the matching values (ones, tens, and hundreds).
Favorable ways we can choose two values are: 3*9*10 = 270C2
Total ways we can arrange 900 students is 900C2.
Please let me know where my logic went wrong if i made a mistake. I'm not 100% sure about this question.
Re: In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
31 Jan 2019, 14:26
kchen1994 wrote:
EgmatQuantExpert wrote:
In a school of 900 students, all of them are coded with a distinct three-digit number from 100 to 999. If two students are select at random, what is the probability that exactly two digits of their codes are same?
A. \(^{81}C_2/^{900}C_2\) B. \(^{90}C_2/^{900}C_2\) C. \(^{162}C_2/^{900}C_2\) D. \(^{243}C_2/^{900}C_2\) E. \(^{270}C_2/^{900}C_2\)
I'm guessing E. Here's my reasoning:
We have a total of 270 numbers that can have matching two digits.
100 -> 200 -> 300 -> 400....-> 900. We have a total of 10 ways we can match each number (00, 11, 22, 33... 99). Our numbers can range from 100 -> 999. This means we have 9 hundreds values we can use. We also have 3 columns that can provide the matching values (ones, tens, and hundreds).
Favorable ways we can choose two values are: 3*9*10 = 270C2
Total ways we can arrange 900 students is 900C2.
Please let me know where my logic went wrong if i made a mistake. I'm not 100% sure about this question.
You are including cases where all three digits are same question mention two digit two be same
Re: In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
Show Tags
18 Feb 2019, 11:03
This question is more about number properties and combinations than probability.
Find numerator using slot method:
1) hundreds digit can be chosen in 9 ways (0 isn't permitted) 2) tens and ones digit can be chosen in 10 ways (0-9) 3) -9 for each repeat of triple digit being same (111, 222 ... 999, since we need EXACTLY 2 same) 4) *3 for the position of the different digit (9*10 - 9) * 3 = 81*3 = 243
gmatclubot
Re: In a class of 100 students, all of them are coded with a distinct ....
[#permalink]
18 Feb 2019, 11:03
close
37% (02:24) correct 
