A die is rolled and a coin is flipped. What is wrong with the following reasoning?:
Let $A=\{1,3,6\}$ (event from die roll), let $B=\{H\}$ (event of heads)
then, $A\cap B=\emptyset$ (since they have no common elements)
and $\Pr(A\cap B)=\Pr(\emptyset)=0$.
I realize that the probability of the intersection of two independent events is the product of their probabilities but I want to know what is 'wrong' with this reasoning.
Best Answer
The problem is the wrong choice of sample space.
The right sample space is $\Omega=\{1,2,3,4,5,6\} \times\{H,T\}$ note that the elementary events are ordered pairs $(\omega_1, \omega_2)$.
Now consider the events
\begin{align*} E_1&=\{(\omega_1, \omega_2) \in \Omega: \omega_1\in\{1,3,6\}\}\\ E_2&=\{(\omega_1, \omega_2) \in \Omega: \omega_2=H\}\,. \end{align*}
You should be able to say what $P(E_1 \cap E_2)$ is, and it's not $0$.
Suppose $P$ is the uniform probability.
$$P(E_1 \cap E_2)=\frac{|E_1 \cap E_2|}{|\Omega|}=\frac{3}{12}=0.25\,.$$