[Math] 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!!!

Answer 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?

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