The events (sum of two dice is $\geq 7$, 2nd die is a 3) are intuitively dependent, but multiplication formula doesn't show that?

Question

If they are dependent, shouldn't the following formula hold?

$$P(A\cap B) = P(A)\cdot P(B\mid A)$$

If I enumerate, $$P(A\cap B) = \{ (4,3); (5,3); (6,3) \} = 3/36 $$

If I use the multiplication rule above, I do not get anything even close to $3/36$, what am I doing wrong?

Thank you.

Answer 1

let $m,n\in \{1,2,3,4,5,6\}$ be the result of first and second die respectively.

Let $A$ be the event that $n=3$, $B$ be the event that $m+n\ge 7$.

$$P(A)=\frac{6\cdot1}{6\cdot 6}=\frac{1}{6}$$ $$P(B\mid A)=P(m+n\ge 7 \mid n=3)$$ $$P(B\mid A)=\frac{3\cdot 6}{6\cdot 6}$$ and the result follows.

Answer 2

Let's count the unbiased outcomes in each event. $$\begin{align}n(\Omega)&= 36\\ n(A\cap B)&=n\{(4,3),(5,3),(6.3)\}\\&= 3\\ n(A)&=n\{(1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),\ldots,(6,6)\}\\&=21\\n(B)&= n\{(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\}\\&=6\end{align}$$

Now, since the outcomes are equally probable, then we may use their relative counts to define the probabilities. $$\begin{align}\mathsf P(A\cap B)&=3/36 &&= n(A\cap B)/n(\Omega)\\\mathsf P(A)&=21/36 &&= n(A)/n(\Omega)\\\mathsf P(B)&=6/36&&= n(B)/n(\Omega)\\\mathsf P(B\mid A) &= 3/21 &&=n(A\cap B)/n(A)\end{align}$$

So the multiplication rule definitely holds, since by definition:$$\mathsf P(A)\mathsf P(B\mid A)~{=(n(A)/n(\Omega))\cdot(n(A\cap B)/n(A))\\ = n(A\cap B)/n(\Omega) \\= \mathsf P(A\cap B)}$$

However, it is clear that $\mathsf P(A\cap B)\neq \mathsf P(A)\mathsf P(B)$ so demonstrating that the events are dependent.