Slips of papers 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. \(\dfrac{31}{1000}\)
B. \(\dfrac{36}{1001}\)
C. \(\dfrac{37}{1001}\)
D. \(\dfrac{37}{1000}\)
E. \(\dfrac{41}{1000}\)
The OA is C
Source: Princeton Review
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From 1000 through 2000 inclusive, there are 1001 integers, so the denominator must be 1001, and we already have only two candidate answers. There are a few ways to compute the numerator. For example, we might separate things into two cases:
- numbers that end in three identical digits, like 1000, 1222, 1333, etc. There are 10 such numbers (nine beginning with '1', because we do not count 1111, and finally the number 2000)
- numbers between 1000 and 1999 which include the digit '1' precisely three times, so numbers like 1711 and 1119. We then have three choices (the hundreds, tens or units) for which digit will not equal 1 , and finally 9 choices for that digit (it can be anything from 0 through 9 besides '1'), for a total of 3*9 = 27 numbers of this type.
So there are 10 + 27 = 37 numbers in the range provided with precisely three identical digits, and the answer is 37/1001.
- numbers that end in three identical digits, like 1000, 1222, 1333, etc. There are 10 such numbers (nine beginning with '1', because we do not count 1111, and finally the number 2000)
- numbers between 1000 and 1999 which include the digit '1' precisely three times, so numbers like 1711 and 1119. We then have three choices (the hundreds, tens or units) for which digit will not equal 1 , and finally 9 choices for that digit (it can be anything from 0 through 9 besides '1'), for a total of 3*9 = 27 numbers of this type.
So there are 10 + 27 = 37 numbers in the range provided with precisely three identical digits, and the answer is 37/1001.
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