Bunuel wrote:
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
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 NumberSub Count = 1; only possible value is 6
Double Digit NumberGroup 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 = 18Three Digit NumberLet'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 = 252Number of digits between 1 and 1000 which have 6 = 252 + 18 + 1= 271
Required probability = 271 / 1000
Option B