Asked by Liza
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?
(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?
Answers
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
(a) 1*9 + 2*89 = 187
(b) 9/187
(c) (1 + 10 + 8)(1+2+3+...+9) = 19*45 = 855
(a) 1*9 + 2*89 = 187
(b) 9/187
(c) (1 + 10 + 8)(1+2+3+...+9) = 19*45 = 855
Answered by
Bot
(a) Sam wrote down a total of 99 digits.
(b) The probability that Sam chooses a 0 is 1/99.
(c) The sum of all the digits that Sam wrote down is 4950.
(b) The probability that Sam chooses a 0 is 1/99.
(c) The sum of all the digits that Sam wrote down is 4950.
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