Slips of paper are numbered from 1000 to 2000. If one paper is selecte

Question

Slips of paper are numbered from 1000 to 2000. If one paper is selected at random, what is the probability of selecting a slip of paper with exactly three identical digits?

(A)  \frac{31}{1000}

(B)  \frac{36}{1001}

(C)  \frac{37}{1001}

(D)  \frac{37}{1000}

(E)  \frac{41}{1000}

Archit3110 · Answer

total pages will be 2000-1000+1 ; 1001

for exactly 3 identical digits starting for series of 1000 to 1999 we get
CASE 1 : starting series from 1 ; 1*1*1*9; 9 such cases as last digit can be any value except 1
now thousandths position is fixed so we have 3!/2! ways to arrange other digits eg 1112 ; 1211; 1121; i.e. 3 ways so for 9 digits it will be 3*9 ; 27 cases

CASE 2 : 
1st digit as 1 and remaining 3 digits are same we get 9 such cases ; eg 1000, 1222,1333,1444 so on

CASE 3 :
1st digit as 2 ; 2000 ; 1 possible

total ways 27+9+1 ; 37

\frac{37}{1001}

option C

ryanstarr · Answer

As this is a probability problem  and  a digits problem, it's likely best solved using a combinatorics-based approach, especially slot method (which tends to work really well for "How many 4-digit numbers have..."-type problems).

First, get the denominator. From 1000 to 2000 inclusive, there are 1001 numbers. You can determine this by calculating the range (2000 - 1000 = 1000) and adding 1 to it (for the number 1000, which we just subtracted away in calculating the range). That already limits our possible answers to B and C, meaning that single step alone gives you a 50% chance of getting this problem correct. In fact, given that 37 shows up in another answer choice as well, C is a pretty safe educated guess. However, if we do want to commit to the rest of the problem, our next task is to figure out how many of the numbers in that range have exactly 3 identical digits. Given that all the numbers in this range start with 1 (except 2000 itself), there are a few ways this can happen:

1 1 1 __, in which the units digit can be one of 9 numbers (0, 2, 3, 4, 5, 6, 7, 8, 9)
1 1 __ 1, in which the tens digit can be one of 9 numbers (the same as above)
1 __ 1 1, in which the hundreds digit can again be one of the same 9 numbers
1 __ __ __, in which all three of the unknown digits can be the same 9 non-1 numbers (1000, 1222, 1333, etc.)

That's 36 total options so far; however, we must add one more: 2000 itself, which has three 0s. As such, we have a total of 37 numbers in that range with three identical digits, and the answer is in fact  C .

Ray007 · Answer

if number coming exactly 3 times is 1
1_ _ _
In this case, exactly 2 places need to be filled by 1 and other place can be anything other than 1

possible values = 9*3

now, the number is anything other than 1, 9 possibilities
and 2000 -->1

therefore total = 27+9+1=37

therefore C

arushi118 · Answer

Can we please add that the number 1000 and 2000 are inclusive.
Also when it is not given if the numbers are inclusive or not -- should we take that as meaning that those numbers are not included?

Slips of paper are numbered from 1000 to 2000. If one paper is selecte

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Slips of paper are numbered from 1000 to 2000. If one paper is selecte

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