# Resolving a counterexample to the most fundamental probability concept; mutually exclusive and independence

Suppose you roll a die $(1-6)$ and toss a coin each once. Let $A$ be the event that I get either heads or tails (let's say tails) and $B$ be the event that I roll a number $(1 - 6)$, (let's say $2$).

So the sample space of $A$ is $\{H, T \}$ and sample space of $B$ is $\{1, 2, 3, 4, 5,6 \}$

Clearly the outcome of $A$ has no effect on $B$, yet $A = \{ T\}$ and $B = \{2\} \implies A \cap B = \phi$

So doesn't this mean I have a contradiction? Since the two events are mutually exclusive and independent? I can't seem to wrap my head over $P(A \cap B) = P(A)P(B)$

EDIT

Wait I am making another mistake again because I found $P(A) = \frac{|A|}{|S|} = \frac{|A|}{|A \times B|} = 1/6$

• Sample space should be $\{H,T\}\times\{1,2,3,4,5,6\}$ – Maesumi Jul 15 '13 at 18:14
• Oh because the tossing and rolling is a single experiement; I forgot about that – Hawk Jul 15 '13 at 18:16
• Just to make sure; the single experiment of the sample space is $S = A \times B$ – Hawk Jul 15 '13 at 18:19
• They are not mutually exclusive: you could toss tails and then roll 2. (This event has probability exactly $\frac12 \frac16 = \frac1{12}$ as expected.) – ShreevatsaR Jul 15 '13 at 19:29
• Independence only makes sense for events defined on the same sample space. – Nate Eldredge Jul 15 '13 at 19:29

## 1 Answer

You should consider elementary events. These consist of a coin toss combined with a roll of the die. Thus your sample space is the direct product $$S = \{H,T\}\times\{1,2,3,4,5,6\}$$

The combined events you describe are projections of these atomic events. When you write $A=\{T\}, B=\{2\}$ then this is only a shorthand notation for

\begin{align*} A &= \{(T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\} \subseteq S \\ B &= \{(H,2), (T,2)\} \subseteq S \end{align*}

Therefore $A\cap B=\{(T,2)\}\neq\emptyset$. You get the probabilities

\begin{align*} P(A) &= \frac{\lvert A\rvert}{\lvert S\rvert} = \frac{6}{12} = \frac12 \\ P(B) &= \frac{\lvert B\rvert}{\lvert S\rvert} = \frac{2}{12} = \frac16 \\ P(A\cap B) &= \frac{\lvert A\cap B\rvert}{\lvert S\rvert} = \frac{1}{12} \end{align*}

• So $A \neq \{H, T \}$ then? – Hawk Jul 15 '13 at 18:25
• Yes, $A = \{(T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\}$ and $B = \{(H,2), (T,2)\}$ making $A\cap B = \{(T,2)\}$ whose probability is, by independence, $P(A)P(B) = \frac{6}{12}\times\frac{2}{12} = \frac{1}{12} = P(T)P(2) = \frac{1}{2}\times\frac{1}{6}$. – Dilip Sarwate Jul 15 '13 at 18:34