If one number is chosen at random from the first 1,000 positive intege

Question

If one number is chosen at random from the first 1,000 positive integers, what is the probability that the number chosen has at least one digit with the number 6?

A. 27/100
B. 271/1000
C. 34/125
D. 44/125
E. 729/1000

gmatophobia · Answer

In this question we have to calculate the number of times 6 will appear in the first 1,000 positive integers.

Among the first 1000 positive integers, a number can be single digit, two digits or three digits

Single Digit Number

Sub Count = 1 ; only possible value is 6

Double Digit Number

Group 1  : The number can end with 6, such as 16, 26, 36 ... 96.

Group Count: 9 numbers.

Group 2  : Alternatively the number can begin with 6_. This is applicable for all numbers between 60 and 69.

So 10 numbers.

Note we have double counted 66 in both Group 1 and Group 2.

Sub count = 10 + 9 - 1 = 18

Three Digit Number

Let's observe between 100 and 199

Group 1  : The number can end with 6, such as 106, 116, 126 ... 196.

So 10 numbers.

Group 2  : Alternatively the number can begin with 6 at its tens position, _6_. This is applicable for all numbers between 160 and 69.

So 10 numbers.

Note : We have double counted 166 (counted in both group 1 and group 2).

So 19 digits.

Now this pattern will repeat for all three digit numbers except for the ones that begin with 6 themselves. So the pattern repeats 8 times.

Hence number of digits 19 * 8 = 152

Group 3  : Numbers starting with 6 (format : 6 _ _ )

Number of digits = 100

Sub Count = 252

Number of digits between 1 and 1000 which have 6 = 252 + 18 + 1= 271

Required probability = 271 / 1000

Option B

PB005 · Answer

P(atleast 1 digit is 6)=1-P(no digit is 6)
total outcomes = 10*10*10 - as 3 spaces can be filled in 10 ways (0-9 any one of these digits)
expected outcome (no digit is 6)= 9*9*9 (0-9 expect 6: any of these digits)
P(no digit is 6) = expected outcome (no digit is 6)/total outcomes
P(atleast 1 digit is 6)=1-P(no digit is 6)
option B

phamthao017 · Answer

P(at least 1 digit is 6)=1- P(no digit is 6)

1-digit number: 8 number with no digit is 6 (except 6)
2-digit number: 90-18=72 number with no digit is 6 (except 18 number with 6: 16...96 and 60...69)
3-digit number: 8.9.9=648 number with no digit is 6 (3-digit number calls abc, so 8 ways to choose a - no 0&6; 9 ways to choose b - no 6; 9 ways to choose c - no 6)
4-digit number: 1 number with no digit 6 (1000)

=> P(at least 1 digit is 6)=1- P(no digit is 6) = 1 - (8+72+648+1)/1000 = 1 - 729/1000 = 271/1000 => B

PSKhore · Answer

Let's take the first 100 numbers: 
In every 10 numbers there's one 6, however starting 60 there are 10 numbers from 60 to 70
So, for each 100
for every 90 numbers there are 9 sixes 
and for the remaining 10 there are 10 sixes.
So, 19 sixes for every 100 numbers.

Now, if we consider 1000, there are 100 sixes starting from 600
So, 9X19 + 100
171+100 = 271
271/1000

OR

Consider a case where no 6 will appear in any of the three digits
So for every digit - there will be 9 numbers : 0,1,2,3,4,5,7,8,9
9*9*9 = 729
Subtract this from 1000
1000 - 729 = 271

If one number is chosen at random from the first 1,000 positive intege

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GMAT Problem Solving (PS) Questions

If one number is chosen at random from the first 1,000 positive intege

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615 to 715 in 15 Days (Ria's Strategies)

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