Suppose a whole number is selected at random between 100 and 999 inclusive.

probability combinatorics permutations combinations

1,069

Solution 1

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?

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

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?

1,069

ChocolateAndMath

Updated on January 25, 2020

Comments

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 = 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!!!
- barak manos almost 7 years
  
  a is wrong. should be $999-100+1=900$.
- wolfies almost 7 years
  
  Less chocolate. More math.
- ChocolateAndMath almost 7 years
  
  Can you explain why?
- Ross Millikan almost 7 years
  
  @ChocolateAndMath You have a fencepost error
- ChocolateAndMath almost 7 years
  
  Googling what a fencepost error is
ChocolateAndMath almost 7 years

Why 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 years

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?)
barak manos almost 7 years

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

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

Let us continue this discussion in chat.