From 000 to 999, probability exactly 1 digit is >5

[Poopsilon]

Homework Statement

If a 3-digit number (000 to 999) is chosen at random, find that probability that exactly 1 digit will be >5

The Attempt at a Solution

So basically I first look at the probability of at least 1 digit being greater than 5, taking into account multiple counting:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C).

Now the part where I'm a bit confused, I need to now subtract the probability of having at least two digits greater than 5. I'm inclined make the subtraction:

-[P(A ∩ B) + P(A ∩ C) + P(B ∩ C) + P(A ∩ B ∩ C)].

But now I'm afraid I've done some multiple counting again and so finally add:

3P(A ∩ B ∩ C)

This comes out to: 3(4/10) - 6(16/100) + 3(64/1000) = .372

Yet the back of the book is saying .432, where have I gone wrong? Thanks.

 

[Dick]

There's a much simpler way to count them than that. Besides, I don't think you even want to count P(A ∪ B ∪ C). You want exactly one of them, not at least one of them. Just count the number of ways you can select which place will hold the large digit. Then multiply by the number of ways to fill in the large digit and the two small digits.

 

[lanedance]

how about this
say first digit >5, we have 4 choices
for the 2nd and 3rd digit <=5, we have 6 choices for each
however cases where the 2nd and 3rd digit are repeated will be counted twice (? possible cases)
Then multiply to account for the cases where it was the 2nd or 3rd digit (only)that was >5

 

[Poopsilon]

Excellent characterizations, much simpler, thanks (insert gender-neutral plural pronoun).

 

[phinds]

Hm ... I must be doing something wrong if .432 is the right answer. I look at it like this:

The chance of a digit being > 5 is 50% and then the chance of the other two being less < 5 at the same time is 50% each. SO ... one combination is 50%x50%x50% = 1/8 and there are 3 ways if it happening, so 3 x 1/8 = 3/8 = .375, not .432

Why is 3/8 not the right answer?

 

[Zoe-b]

Because the chance of a digit being > 5 is 0.4 :P (6,7,8,9).

 

[phinds]

because the chance of a digit being > 5 is 0.4 :p (6,7,8,9).

doh !