# What is intuition behind the product rule of probability and independent events? [duplicate]

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?

• Do these Venn diagrams clarify it? N.B. You quoted the sum rule, not the product rule. Commented Apr 10, 2021 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. Commented Apr 10, 2021 at 19:12
• You have the union and intersection switched in your formula. Commented Apr 10, 2021 at 19:13
• @EricPerkerson, I know, but it is equivalent and necessary to compute the intersection. Commented Apr 10, 2021 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. Commented Apr 10, 2021 at 19:16

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.