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(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 = 999-100-899

b). No. of one's between 200-999 = 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 100-190 = # of one's between 100-119 +(120-190) = 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!!!

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    a is wrong. should be $999-100+1=900$. – 2017-01-24
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    Less chocolate. More math. – 2017-01-24
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    Can you explain why? – 2017-01-24
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    @ChocolateAndMath You have a [fencepost error](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error) – 2017-01-24
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    Googling what a fencepost error is – 2017-01-24

2 Answers 2

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How many outcomes are in $\Omega$?

$999-100+1=900$

What is the probability that the selected number has at least $1$ "1" in it?

$(900-8\cdot9\cdot9)/900=28\%$

What is the probability that the selected number has exactly $2$ "3"’s in it?

$(1\cdot1\cdot9+1\cdot9\cdot1+8\cdot1\cdot1)/900=2.\overline{8}\%$

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    Why are we adding 1? – 2017-01-24
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    @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 $\in[0,5]$? – 2017-01-24
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    @ChocolateAndMath: See [this question](http://math.stackexchange.com/q/879706/131263), and you'll find a few more ways to explain why we need to add $1$. – 2017-01-24
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    Right. So, it is like indexing in for loops. So, it is including 0? So, in the first part I calulcated from 101-999 and not 100-999? (1-9 in place of 0-9?) – 2017-01-24
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    @ChocolateAndMath: Or from $100$ to $998$. – 2017-01-24
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    okay! got it. Can you explain how you got the second part? – 2017-01-24
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52377/discussion-between-chocolateandmath-and-barak-manos). – 2017-01-24
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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?