How do you define a sample space with rigor?

Question

I was reading First Course on Probability by Sheldon Ross and I came across a problem which went like this:
"A customer visiting the suit department of a certain store will purchase a suit with probability $.22$, a shirt with probability $.30$, and a tie with probability $.28$. The customer will purchase both a suit and a shirt with probability $ .11$, both a suit and a tie with probability $.14$, and both a shirt and a tie with probability $.10$. A customer will purchase all $ 3$ items with probability $.06$. What is the probability that a customer purchases (a) none of these items? (b) exactly $1$ of these items?"
Problem a) is easy to solve, what confuses me is part b). Ross solves it in the following way:
The probability that two or more items are purchased is $P(AB ∪ AC ∪ BC) = .11 + .14 + .10 − .06 − .06 − .06 + .06 = .23$ Hence, the probability that exactly $1$ item is purchased is $.51 − .23 = .28.$
Intuitively, I understand why he subtracts the probability of buying two or more things from the probability of buying anything at all. What I do not understand is the rigor behind it. Why am I justified in subtracting one probability from the other?
What I tried to do in order to justify this was saying $P( \mathrm {buying \ 2 \ or \ more \ things)} + P(\mathrm {buying \ 1 \ thing}) + P(\mathrm {buying \ nothing})=1$ since a customer must buy a shirt, or a tie, or a suit, or nothing, therefore the three terms above must add up to the probability of the sample space which equals one.
A sample space is the set of all the outcomes of an experiment, the event space is a set of subsets of the sample space and an event is an element of the event space. I know that $P$ maps events to the unit interval, the only problem I have is how do I define the previous events (namely $\{ \mathrm {buying \ 2 \ or \ more \ things} \}$, $\{ \mathrm {buying \ 1 \ thing} \}$, $\{\mathrm {buying \ nothing} \}$).
If I defined the sample space as $\{\mathrm {shirt, \ suit , \ tie , \ nothing} \}$, then the sets $\{ \mathrm {buying \ 2 \ or \ more \ things} \}$ and $\{ \mathrm {buying \ 1 \ thing} \}$ would not be disjoint and I would not be able to add them together like I did. So if I were asked to write those events explicitly, how would I go about doing it? Am I defining the sample space wrong? I need some clarification on exactly how to define sample spaces such that, if I wanted to, I could write them down explicitly.

Answer 1

If you want to model the number of items which are bought, your sample space could be the set of all tuples $(n_{\mathrm{shirt}},n_{\mathrm{suit}},n_{\mathrm{tie}})$, where the $n$'s are the number of each item bought (each one either $0$ or $1$ if we don't buy more than one of each). Then, for example, the event $$ \{\text{buying 1 thing}\} = \{(1,0,0), (0,1,0), (0,0,1)\}, $$ and the event $$ \{\text{buying 2 things}\} = \{(1,1,0),(1,0,1),(0,1,1)\}. $$

Answer 2

EDIT for clarity: Your logic is correct, the problem here is that an "event" can either mean a singleton set or a set of singleton sets.

So we have the probability space $\Omega = \{suit, hat, tie\}$, but the set of all events is the power set of $\Omega$, i.e. $\mathscr{P}(\Omega) = \{ \emptyset, \{hat\}, \{tie\}, \{suit\}, \{hat, tie\}, \{hat,suit\},\{tie,suit\} \{hat, tie,suit\} \}$.

The reasoning behind subtracting the probability is the inclusion-exclusion principle, for only two events it takes the form $\mathbb{P}(A \cup B) = \mathbb{P}(A)+\mathbb{P}(B) -\mathbb{P}(A\cap B)$, the subtraction of the intersection being in order to avoid double counting.

Also by the way Ross's book is excellent; the only book I would recommend for learning probability for the first time.

EDIT: an explicit example of the inclusion-exclusion principle:

Say we want to calculate the probability of buying a hat and a tie.

Then this event equals: $\{hat,tie\} \cup \{hat, tie,suit\}$, so to calculate the probability $\mathbb{P}(\{hat,tie\} \cup \{hat,tie,suit\} = \mathbb{P}(\{hat,tie\})+\mathbb{P}(\{hat,tie,suit\}) - \mathbb{P}(\{hat,tie\})$. This is because $\{hat,tie\} \subset \{hat,tie,suit\}$ i.e. $\{hat,tie\} \cup \{hat, tie,suit\}=\{hat,tie,suit\}$, so if we had included the probability of $\{hat,tie\}$ we would have been double-counting.

This isn't the best example admittedly; Feller vol. 1 or Shiryaev explain this better than I do.