I have no idea here

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?

Guest Apr 17, 2023

#1**+3 **

a) Let us first find out the total number of numbers. There are 99 as it is a pattern going up by 1s starting on 1.

To solve this problem, the first thing I did was to separate the 1-digit and 2-digit numbers. We know that the only 1-digit numbers are 1-9, and everything else is 2-digit numbers.

Let us list this information:

1-digit = 9x

2-digit = 90x (99-9)

Now to find the total number of digits, we can multiply the number of x-digit numbers by x.

\(9\cdot1=9\) (1-digit total)

\(90\cdot2=180\)(2-digit total)

\(9+180=189\)(total)

Sam wrote down 189 digits in total.

b) For this one, to find the probability of getting a 0 out of all digits, we can divide the number of 0s by the total number of digits.

As there are only a few 0s, we can list them out:

10,

20,

30,

40,

50,

60,

70,

80, and

90.

In total, there are 9 0s.

As we have already evaluated the total number of digits in part a), we can divide 9 by 189.

The probability that Sam chooses a 0 is 9 out of 189. (I think you should be able to do some simplifying yourself???)

c) I am so sorry, I'll come back to this question in about 20 mins, but I don't have time right now to explain it in detail.

However, I can tell you that it has something to do with this:

\(1+99\),

\(2+98\),

\(3+97\)...

\(49+51\),

\(50\)

I hope you will be able to solve it from here, but if not, I'll be back.

Hope this helped,

BlackjackEd Apr 17, 2023