I'm very conufsed by this problem. Can any assist?

Sam writes down the numbers 1, 2, 3, ..., 99

(a) How many digits did Sam write, in total?

(b) Sam chooses one of the digits written down, at random. What is the probability that Sam chooses a 0?

(c) What is the sum of all the digits that Sam wrote down?

sandwich Feb 12, 2023

#1**0 **

(a) Sam writes down 99 numbers, which means he writes down a total of 99 digits.

(b) Sam writes down 10 zeros (0, 10, 20, ..., 90), so the probability of choosing a zero is 10/99, or about 0.101010101.

(c) The sum of all the digits from 1 to 99 is (1 + 2 + ... + 9) + (10 + 11 + ... + 19) + ... + (90 + 91 + ... + 99) = 45 * 10 + sum of all digits from 1 to 9. The sum of all digits from 1 to 9 is 45, so the total sum is 45 * 10 + 45 = 450 + 45 = 495.

Mathefreaker2021 Feb 12, 2023