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?
+79 (a)
Let's look at a simplified version: 1, 2, 3, 4. How many numbers are in that?
Obviously 4. That's the last number in the sequence.
Therefore, using the same logic, Sam wrote down 99 digits.
+79 (b)
He didn't write down a 0. Using \(\text{Probability}=\dfrac{\text{number of successful outcomes}}{\text{total possible outcomes}}\), the probability is 0.
+79 (c)
The sum of n natural numbers is \(n(n+1)\over 2\), so we have \(\dfrac{99(100)}{2}=4950\)
