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?
+32 (a) 99 digits in total
(b) 0 because Sam didn't write the number 0
(c) 4950
Hope this helps ^^
a) Sam has total digits of 189
b) The possibility is 9/189; as the numbers 10, 20, 30 ...90 have the number 0.
c) The sum of n = n(n+1)/ 2
-> 99(99 + 1) / 2
-> 99(100) / 2
-> 9900 / 2
-> 4950
