What is intuition behind the product rule of probability and independent events?

Question

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. — 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. — jck21, Apr 10 '21 at 19:12
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. — jck21, 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
@joe You've updated the post to include numbers; thanks! The events are not independent, 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? — jck21, 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.) 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:) — jck21, 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

score 1 · Accepted Answer · answered Apr 10 '21 at 20:05

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.

What is intuition behind the product rule of probability and independent events?

1 Answers1