# Probability of a tetrahedron (die with 4 faces)?

I have been doing some questions from an exam review with no solution and I have no idea how to work this problem. I know that $Pr(A_1) = \frac{1}{2}$, $Pr(A_2) = \frac{1}{2}$, $Pr(A_3) = \frac{1}{2}$, $Pr(A_1 \cap A_2) = \frac{1}{4}$, $Pr(A_1 \cap A_3) = \frac{1}{4}$, $Pr(A_2 \cap A_3) = \frac{1}{4}$ but how do I use this information?

Let's look at a tetrahedron (die with 4 faces). Each one of these faces has one of the bitstrings "110", "101", "011", "000".

For k=1,2,3, define the event

Ak = "the bitstring written on the bottom face has 0 (zero) at position k" For example, if the bitstring at the bottom face is 101, then A1 is false, A2 is true, and A3 is false.

• en.wikipedia.org/wiki/Pairwise_independence – Alex Mar 20 '14 at 17:04
• In this link there's a link to the definition of mutual independence. This fully solves your question. – Alex Mar 20 '14 at 17:08
• @Alex thats soo strange the example in your link is almost exactly the same question thanks – James Mar 20 '14 at 17:09
• you are welcome – Alex Mar 20 '14 at 17:12
• Essentially a duplicate of this question which is already on hold. – Dilip Sarwate Mar 20 '14 at 17:12

Two events are independent exactly if $$\mathbb{P}(X \cap Y) = \mathbb{P}(X) \cdot \mathbb{P}(Y) \text{.}$$
Since per your question, $\mathbb{P}(A_1) = \mathbb{P}(A_2) = \frac{1}{2}$ and $\mathbb{P}(A_1 \cap A_2) = \frac{1}{4}$, the requirement above is satisfied (because indeed $\frac{1}{2}\frac{1}{2} = \frac{1}{4}$), and therefore $A_1$ and $A_2$ are independent.
For mutual independence, you need for events $E_1,\ldots,E_n$ that $$\mathbb{P}(X_1 \cap X_2 \cap \ldots X_m) = \mathbb{P}(X_1) \cdot \mathbb{P}(X_2) \cdot \ldots \cdot \mathbb{P}(X_m)$$ for every selection of events $X_1,\ldots,X_m$ from $E_1,\ldots,E_n$.