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805+ Level|   Probability|      
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In a class of 100 students, all of them are coded with a distinct .... Updated on: 30 Jan 2019, 11:46
Originally posted by EgmatQuantExpert on 30 Jan 2019, 01:52.
Last edited by EgmatQuantExpert on 30 Jan 2019, 11:46, edited 3 times in total.
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95% (hard)

Question Stats:

39% (02:33) correct 61% (02:21) wrong based on 188 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. \(^{270}C_2/^{900}C_2\)


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D
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In a class of 100 students, all of them are coded with a distinct .... 01 Feb 2019, 02:02
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Solution


Given:
    • Number of students in a school = 900
    • All of them are coded with a distinct number from 100 to 999
    • Two students are selected at random

To find:
    • Probability that exactly two digits of their codes are same

Approach and Working:
    • Total number of ways of selecting 2 students from a total of 900 students = \(^{900}C_2\)

Now, let’s find the number of 3-digit numbers in which exactly two digits are same
Case 1: Units digit = tens digit
    • Units and Tens digit can take any number from 0 to 9, and the hundreds digit can take any value from 1 to 9
      o So, total such numbers = 10 * 9 = 90

    • But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
      o Total such numbers are {111, 222, 333, …, 999} = 9

Case 2: Hundreds digit = tens digit
    • Hundreds and tens digit can take any number from 1 to 9, and units digit can take any value from 0 to 9
      o So, total such numbers = 10 * 9 = 90

    • But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
      o Total such numbers are {111, 222, 333, …, 999} = 9

Thus, total numbers in this case = 90 – 9 = 81

Case 3: Units digit = hundreds digit
    • Units and hundreds digit can take any number from 1 to 9, and the tens digit can take any value from 0 to 9
      o So, total such numbers = 10 * 9 = 90

    • But, this includes numbers with all three digits same. So, we need to eliminate such numbers.
      o Total such numbers are {111, 222, 333, …, 999} = 9

So, total such numbers = 81 + 81 + 81 = 243
    o We can select two numbers from this in \(^{243}C_2\) ways

Therefore probability = \(\frac{^{243}C_2}{^{900}C_2}\)

Hence the correct answer is Option D.

Answer: D

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In a class of 100 students, all of them are coded with a distinct .... 30 Jan 2019, 05:01
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EgmatQuantExpert
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 .... Updated on: 31 Jan 2019, 02:06
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.
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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
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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In a class of 100 students, all of them are coded with a distinct .... 30 Jan 2019, 11:49
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Hi,

There was a mistake in the question statement. We have corrected it now.

Regards,
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In a class of 100 students, all of them are coded with a distinct .... 30 Jan 2019, 20:46
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EgmatQuantExpert
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.
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In a class of 100 students, all of them are coded with a distinct .... 31 Jan 2019, 00:40
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i am getting 234c2/900c2

the 2 repeatative digit can be selected in 9ways and single digit must be selected apart from repeatative digit so 8ways
thus can be arranged in3!/2!

so total arrangement is 8*9*3

no considering 0 as one digit

it can be like 101,...909 so 9 ways

when there are two zeros

number will be 100......900 again 9 ways

so total count is 8*9*3+9+9=234

234c2/900c2

help where i went wrong in counting
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In a class of 100 students, all of them are coded with a distinct .... 31 Jan 2019, 14:26
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kchen1994
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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. \(^{270}C_2/^{900}C_2\)

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

u sud exclude 111,222

hope that way
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In a class of 100 students, all of them are coded with a distinct .... 18 Feb 2019, 11:03
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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
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In a class of 100 students, all of them are coded with a distinct .... 27 Mar 2020, 23:45
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100-900

Three slots --> 1-9, 0-9, 0-9
Total combos numbers = 9*10*10 = 900

Scenario 1: first digit and 2nd or 3rd digit match
(9*1*9) * 2! = 162 --> nine numbers for the first choice, pick the same on the 2nd, and don't pick the same on the 3rd. Multiply by 2! to get combos of 1-2 match or 1-3 match

Scenario 2: 2nd and 3rd digit match WITH ZERO
(9*1*1) = 9

Scenario 3: 2nd and 3rd digit match WITHOUT ZERO
(9*8*1) = 72 --> 2nd becomes 8 because no zero and no match with the first number

162+9+72 = 243

243 total combos of two being the same

243C2/900C2
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In a class of 100 students, all of them are coded with a distinct .... 26 Feb 2021, 11:13
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The question is not clear. I thought we have to pick two numbers such that exactly two digits are same between those two numbers:
i.e.
123, 314
in this case, 1 and 3 from first number matches 3 and 1 in second number.

But the question is about picking two numbers such that each number has exactly two digits matching within itself. This clarification may disambiguate the question
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In a class of 100 students, all of them are coded with a distinct .... 06 May 2022, 22:27
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