Question

An integer from 100 to 999, inclusive is to be chosen at random. What is the probability that the no. chosen will have 0 as at least 1 digit

Answer 1

$\mathsf{\huge{Answer}}$

First we have to find the total number of integers existing between 100 and 999

= 999 - 100 + 1 = 900

There are three possibilities,

0 in 3 digits : 0

0 in 2 digits : 100, 200, 300, 400, 500, 600, 700, 800, 900 = 9

0 in 1 digit : 101 - 110, 120, 130, 140, 150, 160, 170, 180, 190 = 180.

We can go on like this,

901 - 910, 920, 930, 940, 950, 960, 970, 980, 990 : 18

$Or\:18\times9 = 162.$

Now, 0+9+162 = 171

So, the probability is = $\frac{No.\:digits\:with\:0\:as\:at\:least\:one\:digit}{Total\:no.\:of\:digits}$

= $\frac{171}{900}$

= $\frac{19}{100}$

= 0.19