Question

An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit.
A. 19/900
B. 81/900
C. 90/900
D. 171/900
E. 271/1000

Please try and answer the question, I will not give you the points if you don't try to answer it correctly.

Answer 1

Total number of digits series = 100, 101, 102 ...... 999

so, a(n) = a + (n - 1)d
999 = 100 + (n - 1)1
999 - 100 = n - 1
899 + 1 = n
n = 900

Now, Numbers with 0 are: 9 sets of (100, 101, 102 .... 110) (11 number in each set)

Then, 9 sets of numbers like: (120,  130, 140 .... 190) (8 number in each set)

So, numbers with 0 as a digit = 11 * 9 + 9 * 8 = 171

Now, we know Probability = Favorable Outcomes / Total Outcomes
P = 171 / 900

In short, Your Answer would be Option D

Hope this helps!