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.

  • $\begingroup$ The sample space here is presumably the power set of $\{\mathrm {shirt, \ suit , \ tie } \}$ and has $8$ possibilities $\endgroup$
    – Henry
    Apr 12, 2016 at 23:25

2 Answers 2


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 $$ \{\text{buying 2 things}\} = \{(1,1,0),(1,0,1),(0,1,1)\}. $$


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.

  • $\begingroup$ But in the exclusion inclusion principle, what intersections and unions am I calculating? $\endgroup$
    – GuPe
    Apr 13, 2016 at 2:35
  • $\begingroup$ Well it depends on what probability you want to calculate. I edited my answer to give an explicit example. $\endgroup$ Apr 13, 2016 at 13:36
  • $\begingroup$ I fail to understand the phrase "an "event" can either mean a single set or a set of sets." Every event is a subset of Omega and the elements of Omega could be sets themselves in which case events would be sets of sets... But this fact is universally true, so why underline it as if it was crucial or explaining anything, especially since the examples you give afterwards are not of this nature? $\endgroup$
    – Did
    Apr 13, 2016 at 17:20
  • 1
    $\begingroup$ This would enrich the power set with many additional composite events -let's leave this as an exercise for the OP! $\endgroup$ Apr 13, 2016 at 20:12
  • 1
    $\begingroup$ "the problem here is that an "event" can either mean a singleton set or a set of singleton sets." Again, this is awfully confusing. Are we supposed to understand that, for example, if Ω={suit,hat,tie} then {{suit},{hat}} is an event? If we are not (as I hope, since {{suit},{hat}} is not an event, at least when suit, hat, and tie are three different objects), what use is the sentence? $\endgroup$
    – Did
    Apr 13, 2016 at 20:23

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