I will present two problems alongside solutions, student is doing problems of type I like a cakewalk but has several issues with the problems of type 2;

Type I

Consider an experiment of rolling two dice:

Sample space $$ S = \{(1,1),(1,2),(1,3), \cdots, (6,6) \} $$

Let $A$ be the event of getting 6 as sum on two dice:

Event $$A = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$

Let $B$ be the event of getting 4 on first die:

Event $$B = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$$

Now, the probability of getting sum of 6 on two dice given that first die appears as 4 is given by

$$p(A \mid B) = \dfrac{p(A\cap B)}{p(B)} = \dfrac{p\{(4,2)\}}{p(B)} = \dfrac{1}{6}$$

Type II

Let $C$ be the event of having cancer and $p(C) = 1/2$

The probability of having both tumor and cancer is $p(C\cap T) = 1/6$

Then the probability of having tumor given cancer is given by

$$p(T\mid C) = \dfrac{p(T\cap C)}{p(C)} = \dfrac{1}{3}$$

Students are understanding type I problems and type 2 problems well, but some students are asking to present $C, T, C\cap T$ in terms of sets. I tried to convince them using Venn diagrams. But they are asking for either roster form (for discrete sets) or set builder form(for any set).

I am trying to do like follows

\begin{align} \text{Sample space} = S &= \{x \mid \text{$x$ is a living being} \} \\ C &= \{x \mid \text{$x$ has cancer}\} \\ C\cap T &= \{x \mid \text{$x$ has both tumor and cancer}\} \\ \end{align}

Is it right way to do? Students are facing difficulty and they are asking every event inform of set, since definition of event is that it is a subset of sample space (which is a set).

  • 1
    $\begingroup$ I might say $S = \{ x \mid \text{$x$ is a person} \}$ (since I am guessing that your sample space is people, not all living things (can a paramecium even get cancer?), and I would likely define the set $T := \{ x \mid \text{$x$ has a tumor} \} \subseteq S$, but it otherwise looks fine. That being said "$T$ is the set of people with tumors" is a reasonable definition of set in this context; I wonder why you are getting so much guff about it... $\endgroup$
    – Xander Henderson
    Commented Jan 15, 2018 at 15:33
  • 1
    $\begingroup$ I would be inclined to write $$ S = \{ x \mid x \text{ is a sample} \} \qquad \qquad C = \{ x \mid x \text{ is in event $C$} \}$$ and so forth if someone insisted on writing the events as sets in a context where they don't matter. $\endgroup$
    – user797
    Commented Jan 15, 2018 at 20:17
  • 1
    $\begingroup$ There's a step missing in both solutions, and this serves to hide the fact that in the Type I solution one is likely answering by simply substituting frequencies (not probabilities). That is, it submarines the possibility that one thinks $p((4, 2)) = 1$ and $p(B) = 6$; this in turn can then feed errors in Type II. $\endgroup$ Commented Jan 16, 2018 at 2:12
  • $\begingroup$ The tag primary-education is certainly wrong here, no? $\endgroup$
    – mweiss
    Commented Jan 21, 2018 at 3:55
  • $\begingroup$ @mweissThank you $\endgroup$
    – hanugm
    Commented Jan 21, 2018 at 7:13

3 Answers 3


To use set builder notation you need to carefully state to which set every element you want to discuss belongs. In general it goes

$P = \{x\in\text{some set} : x\text{ satisfies some truth evaluable proposition}\}$.

Here's how I would do it in your case:

Let $P$ be the set of all humans, let $C=\{x\in P: x\text{ has cancer}\}$ and let $T=\{x\in P: x\text{ has a tumor}\}$. Then $C\cap P$ is the set of humans that both have cancer and a tumor, or $C\cap P = \{x\in P: (x\text{ has a tumor})\wedge (x\text{ has cancer})\}$. In the previous, the wedge notation is the logical 'and'.


The question is very confusingly stated. (Hopefully it is more clearly done in native language.) I can't help with the logic barrier for the students, but if they are having a hard time with the logic and then you add on confusing language, this makes things worse. Or may be more the problem than the logic barrier itself.

"Let T be the event of having cancer and p(C)=1/2".

Is T tumor or cancer??? And why have you combined such separate ideas (definition of T tumor and probability of C cancer) into one joined compound sentence? [You are lacking a comma, too. Wouldn't normally nit about this--make mistakes myself--but it adds to the bafflement.] For that matter what does "cancer" and "tumor" mean? (Note how I add the term "detectable" below to help with the huh factor of people equating C and T.)

Let me try:

  1. C is the fraction of the population that has cancer.
  2. T is the fraction with a detectable tumor.*


A. Probability of cancer C in the population is 0.5 B. Probability of having both cancer C and a detectable tumor T is 1/6.

Question: What is the probability of having a detectable tumor T, given having cancer C?

Answer: (your equation)

P.s. This is just a student math problem but mistakes in the medical literature are easy to make and common. In some cases even affecting patient treatment. So try to be super precise. I'm not perfect either probably have some mistakes myself. But I do know the importance of clear communication in medical work!

*Not clear to me if you are considering benign detectable tumors (i.e. possible to have T sans C). This has implications to how to think about the problem in real world. For instance, you could have a person who has C (cancer) and has T (detectable tumor) but the detected tumor is benign. Thus examination, biopsy, surgery etc. might not detect or treat the person's cancer. I don't think this affects your math problem, but just be careful.

  • $\begingroup$ True, sorry for that $\endgroup$
    – hanugm
    Commented Jan 24, 2018 at 5:02

Since they can do Type-1 problems, I'd begin by trying to make your Type-2 problem look as similar to Type 1 as possible, despite the fact that, in the Type-2 problem, we don't actually know the size of the sample space. So I'd temporarily assume some specific size, say $6$ (to make the arithmetic easy). So we have $6$ people of whom $3$ have cancer, and one of those $3$ also has a tumor. Let them work out the conditional probability in this case, which I hope they can do because this version of the question is very similar to Type 1. Then, re-do the problem with a sample space of size, say, 24. And again with 60 or 600, etc., until it becomes clear to them that the information in the problem is enough to determine the answer, regardless of the size of the sample space. The formula for conditional probability in terms of absolute probabilities summarizes the results of these calculations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.