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In a class of 100 students, all of them are coded with a distinct ....
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Updated on: 30 Jan 2019, 10:46
3
00:00
A
B
C
D
E
Difficulty:
85% (hard)
Question Stats:
38% (02:19) correct 62% (02:11) wrong based on 37 sessions
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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 ....
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30 Jan 2019, 04:01
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
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In a class of 100 students, all of them are coded with a distinct ....
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Updated on: 31 Jan 2019, 01:06
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. \(^{270}C_2/^{100}C_2\)
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Originally posted by Archit3110 on 30 Jan 2019, 05:58.
Last edited by Archit3110 on 31 Jan 2019, 01:06, edited 1 time in total.
Re: In a class of 100 students, all of them are coded with a distinct ....
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30 Jan 2019, 19:46
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 ....
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31 Jan 2019, 13:26
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 ....
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18 Feb 2019, 10:03
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 ....
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18 Feb 2019, 10:03
close
38% (02:19) correct 
