Suppose a whole number is selected at random between 100 and 999 inclusive.
Solution 1
How many outcomes are in $\Omega$?
$999100+1=900$
What is the probability that the selected number has at least $1$ "1" in it?
$(9008\cdot9\cdot9)/900=28\%$
What is the probability that the selected number has exactly $2$ "3"’s in it?
$(1\cdot1\cdot9+1\cdot9\cdot1+8\cdot1\cdot1)/900=2.\overline{8}\%$
Solution 2
For b you shouldn't count the number of ones because you will double count numbers like $112$ that have two of them. Once you have the correct answer for a it is easiest to count and subtract the numbers that do not have $1$s. How many choices for the hundreds digit if $1$ is prohibited? The other digits? How many numbers have no $1$s?
For c, how many ways to select which two digits are $3$? How many ways to select the third digit?
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ChocolateAndMath
Updated on January 25, 2020Comments

ChocolateAndMath almost 4 years
(a) How many outcomes are in Ω?
(b) What is the probability the selected number has at least one 1 in it?
(c) What is the probability the selected number has exactly two 3’s in it?
What I have so far:
a). Sample Space = 999100899
b). No. of one's between 200999 = 8 rows *10 + 8 times 10(201, 211, 212,.. 301,311,..991) + 8 times 8(for the extra 1's like 211, 311..911) = 704 No. of one's between 100190 = # of one's between 100119 +(120190) = 11+17 + (8*10 + (8)) = 28 + 88 = 116
Total no.of one's in it = 116 + 704 = 820 probability of getting atleast one 1 in it = 820/ 899
So, I don't think I'm approaching the problem right at all. My answer looks very wrong. Can someone help me to come on the right track?
I have no idea on how to do the third part with a different strategy than I have used so far. Any help would be much appreciated!!!

barak manos almost 7 yearsa is wrong. should be $999100+1=900$.

wolfies almost 7 yearsLess chocolate. More math.

ChocolateAndMath almost 7 yearsCan you explain why?

Ross Millikan almost 7 years@ChocolateAndMath You have a fencepost error

ChocolateAndMath almost 7 yearsGoogling what a fencepost error is


ChocolateAndMath almost 7 yearsWhy are we adding 1?

barak manos almost 7 years@ChocolateAndMath: The analogy is "$1$ rope has $2$ ends". Try it with a smaller range, and you'll understand. For example, how many values are $\in[0,5]$?

barak manos almost 7 years@ChocolateAndMath: See this question, and you'll find a few more ways to explain why we need to add $1$.

ChocolateAndMath almost 7 yearsRight. So, it is like indexing in for loops. So, it is including 0? So, in the first part I calulcated from 101999 and not 100999? (19 in place of 09?)

barak manos almost 7 years@ChocolateAndMath: Or from $100$ to $998$.

ChocolateAndMath almost 7 yearsokay! got it. Can you explain how you got the second part?

ChocolateAndMath almost 7 yearsLet us continue this discussion in chat.