I just bumped into a simple question. Let's say I want to compute the probability of taking both Math and Science courses (i.e., $P(M \cap S)$) given this information:

Total class size is 10;
7 students take Math and 5 students take Science.
Only one student takes neither of them. What is the probability that a student takes both Math and Science?

Then I know

$$ \begin{align} P(M \cap S) &= P(M)+F(S)-P(M \cup S) \\ &=0.7+0.5-0.9 \\ &=0.3 \end{align} $$

(easily derived from a Venn Diagram)

but wonder why I can't do $P(M\cap S)=P(M)\times P(S)=0.7\times 0.5=0.35$ in this case, even if $M$ and $S$ seem to be independent events but the result is different). What's the intuition behind the product rule, and why are the answers different?

  • $\begingroup$ Do these Venn diagrams clarify it? N.B. You quoted the sum rule, not the product rule. $\endgroup$ Commented Apr 10, 2021 at 19:06
  • $\begingroup$ @AryaMcCarthy, Yes, Venn diagram clarifies it. It's for sure. But my question is why I can't simply multiply individual probability even if they are independent events. $\endgroup$
    – jck21
    Commented Apr 10, 2021 at 19:12
  • 1
    $\begingroup$ You have the union and intersection switched in your formula. $\endgroup$ Commented Apr 10, 2021 at 19:13
  • $\begingroup$ @EricPerkerson, I know, but it is equivalent and necessary to compute the intersection. $\endgroup$
    – jck21
    Commented Apr 10, 2021 at 19:15
  • $\begingroup$ @EricPerkerson It's not in the usual format, but it's still correct. Try adding $P(M \cup S)$ to both sides and subtracting $P(M \cap S)$ from both sides. $\endgroup$ Commented Apr 10, 2021 at 19:16

1 Answer 1


Turning my comments into an answer:

These Venn diagrams are a good starting point for understanding how the probabilities of two events interrelate.

If two events are indeed independent, then you can compute $P(A \cap B) = P(A) \times P(B)$. In fact, independence is defined in terms of these joint and marginal probabilities.

In the example you've shown, they're not independent: $M$ and $S$ affect each other. If they were independent, $P(M \mid S)$ would equal $P(M \mid \neg S)$ and $P(M)$. It wouldn't matter what value $S$ took when you consider $M$. But looking at the data, $P(M \mid S) = 3/5$, $P(M \mid \neg S) = 4/5$, and $P(M) = 7/10$. Knowing information about $S$ affects your knowledge of $M$, and vice versa.


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