Question

An integer from 300 through 780, inclusive is to be chosen at random. Find the probability that the number is chosen will have 1 as at least one digit.

(a) \(\frac{171}{900}\)

(b) \(\frac{43}{860}\)

(c) \(\frac{231}{546}\)

(d) \(\frac{31}{701}\)

I got this question at a job interview.

I'm obligated to ask this question of Discrete Probability topic in chapter Discrete Probability of Discrete Mathematics

Answer 1

The correct answer is (a) \(\frac{171}{900}\)

The best explanation: The number of numbers that don’t have one anywhere 9^3 = 729 is (9 possibilities for each individual digit), and there are 9*10^2 = 900 numbers overall (9 possibilities for hundreds, 10 for the tens and units), so there are 900 – 729 = 171 numbers with at least a one and thus \(\frac{171}{900}\) probability.