1. The problem statement, all variables and given/known data If a 3-digit number (000 to 999) is chosen at random, find that probability that exactly 1 digit will be >5 3. 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.