I was given a problem in my stats homework that I had some trouble with. The question is as follows: If three six-sided dice are thrown, what is the probability of obtaining a four when obtaining a four is an event in which any of the following is true:
- At least one die is a 4
- Any two dice sum to 4
- The sum of all three dice is 4
I solved the problem as follows: We are solving for $$P(1 \cup2 \cup 3) = P(1) + P(2) + P(3) - P(1 \cap 2)$$ $P(1) = \frac{91}{216}$ and $P(3) = \frac{3}{216}$ are straightforwards. To solve for $P(2)$, I did the following: $$P(2) = \frac{3}{36} * (3C2) - \frac{8}{216} = \frac{46}{216}$$ where $\frac{8}{216}$ is subtracting duplicates (I obtained by just counting). Finally, I get $P(1 \cap 2)$ by counting as well with: $$P(1 \cap 2) = 9/216$$ This gives me: $$P(1 \cup 2 \cup 3) = \frac{91 + 3 + 46 - 9}{216} = \frac{131}{216}$$ I confirmed this answer with some Haskell code and it seems to be correct.
However, I'm not particularly satisfied with my approach as it is error prone to have to count the probability of certain events. Can anyone provide a more elegant solution to this problem?
Good Question
Answer: $$\newcommand{\c}[2]{\binom{#1}{#2}} \newcommand{\u}{\underbrace} \newcommand{\t}{\text} \newcommand{\b}[1]{\left(#1\right)} \newcommand{\f}{\frac} \begin{align} P(A\cup B\cup C)&=P(A)+P(B)+P(C)-P(AB)-P(BC)-P(CA)+P(ABC)\\ P(A)&=1-(5/6)^3=91/216\\ P(B)&=\f1{6^3}\b{\u{\b{\u4_{2,4,5,6,}\cdot3!+\u2_{3,1}\cdot3}}_{1,3}+\u{\b{\u5_{1,3,4,5,6}\cdot3+\u1_2}}_{2,2}}=46/216\\ P(C)&=3/216\\ P(AB)&=\f1{216}\binom32\binom31=9/216\\ P(BC&=P(CA)=P(ABC)=0 \end{align}$$