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

Question

(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!!!

a is wrong. should be $999-100+1=900$. – barak manos Jan 24 '17 at 16:54 — barak manos, Jan 24 '17 at 16:54
Can you explain why? – ChocolateAndMath Jan 24 '17 at 16:56 — ChocolateAndMath, Jan 24 '17 at 16:56
@ChocolateAndMath You have a fencepost error – Ross Millikan Jan 24 '17 at 16:58 — Ross Millikan, Jan 24 '17 at 16:58
Googling what a fencepost error is – ChocolateAndMath Jan 24 '17 at 17:01 — ChocolateAndMath, Jan 24 '17 at 17:01

score 1 · Answer 1 · answered Jan 24 '17 at 16:57

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?

score 1 · Accepted Answer · answered Jan 24 '17 at 17:00

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?

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

answered Jan 24 '17 at 17:00

barak manos

43,109

Why are we adding 1? – ChocolateAndMath Jan 24 '17 at 17:04
@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 Jan 24 '17 at 17:06
@ChocolateAndMath: See this question, and you'll find a few more ways to explain why we need to add $1$. – barak manos Jan 24 '17 at 17:08
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?) – ChocolateAndMath Jan 24 '17 at 17:09
@ChocolateAndMath: Or from $100$ to $998$. – barak manos Jan 24 '17 at 17:09
okay! got it. Can you explain how you got the second part? – ChocolateAndMath Jan 24 '17 at 17:11
Let us continue this discussion in chat. – ChocolateAndMath Jan 24 '17 at 17:12

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

2 Answers2