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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?

kjetil b halvorsen
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joe
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  • Do [these Venn diagrams](https://brilliant.org/wiki/probability-rule-of-sum/) clarify it? N.B. You quoted the sum rule, not the product rule. – Arya McCarthy Apr 10 '21 at 19:06
  • @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. – joe Apr 10 '21 at 19:12
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    You have the union and intersection switched in your formula. – Eric Perkerson Apr 10 '21 at 19:13
  • @EricPerkerson, I know, but it is equivalent and necessary to compute the intersection. – joe Apr 10 '21 at 19:15
  • @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. – Arya McCarthy Apr 10 '21 at 19:16
  • @Joe If they're indeed independent, then you _can_ do that. Please edit the question to include an example where the numbers didn't work out. – Arya McCarthy Apr 10 '21 at 19:20
  • Oh, you're right, my mistake. – Eric Perkerson Apr 10 '21 at 19:21
  • @AryaMcCarthy, sure I will do it right away. – joe Apr 10 '21 at 19:23
  • @joe You've updated the post to include numbers; thanks! The events are [not independent](https://online.stat.psu.edu/stat800/lesson/how-do-we-check-independence), which is why you can't simply multiply probabilities. – Arya McCarthy Apr 10 '21 at 19:38
  • @AryaMcCarthy, thanks Arya, my understanding is that taking math and science do not affect or interfere with each other, meaning *independence*. Could you explain why they are not independent? – joe Apr 10 '21 at 19:41
  • They do seem to affect each other, based on the numbers you've shown. ([Independence is _defined_ in terms of the joint and marginal probabilities](https://online.stat.psu.edu/stat800/lesson/how-do-we-check-independence).) 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 $P(M) = 7/10$. So knowing information about $S$ affects your knowledge of $M$, and vice versa. – Arya McCarthy Apr 10 '21 at 19:49
  • @AryaMcCarthy, thanks for super simple and clarifying explanation. It makes a perfect sense:) – joe Apr 10 '21 at 19:57
  • See also: https://stats.stackexchange.com/questions/517297/random-sampling-and-independence-in-a-real-world-problem/517317#517317 – Peter O. Apr 10 '21 at 21:41

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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.

Arya McCarthy
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