In a box, we have $5$ dice. Three of them are proper, one has two sixes, and a second one has six sixes.
We’re given the events
- $A$: after the first throw, we have a six.
- $B$: after the next two throws, we have six and something else.
We want to find the probability $$P(A \cap B \mid \text{proper dice}).$$
My attempt is: $$ (1/6)\times(1/6)\times(5/6)=\left(5/6^3\right).$$ I know that this is the wrong answer. The question is, why?
General speaking for two events $A$ and $C$: $$\mathbb{P}(A \mid C) = \frac{\mathbb{P}(A \cap C)}{\mathbb{P}(C)}.$$
So you calculated $\mathbb{P}(A \cap B \cap \textrm{ proper dice})$, so you must still divide by the probability of picking a proper dice.
To be more precise, you need to consider your event space $\Omega$, which can be given as $$ \Omega = \{1, 2,3,4,5,6\}^3 \times \{1,2,3,4,5\}.$$ Here the tuple $(x,y,z,a)$ means: first roll is $x$, second roll is $y$, third roll is $z$ and the dice is number $a$.
Now, you need to find a probability function on this space, which is given by $$\mathbb{P}((x,y,z,a)) =\left\{ \begin{matrix} \frac{1}{5\cdot 6^3} & a \in \{1,2,3\} \\ \frac{1}{5\cdot 6^3} & a=4 \textrm{ and } x,y,z\in\{1,2,3,4\} \\ \frac{2}{5\cdot6^3} &a=4 \textrm{ and only one of }x,y,z \textrm{ is } 6 \\ \frac{4}{5 \cdot 6^3} & a =4 \textrm{ and only two of }x,y,z \textrm{ are } 6 \\ \frac{8}{5\cdot6^3} & a=4 \textrm{ and } x=y=z=6 \\\frac{1}{5} &a=5 \textrm{ and } x=y=z=6\end{matrix}\right.$$
Now, what is the event of first a $6$ and then exactly one $6$ in the last two throws and a fair dice? That is $\{(6,y,z,a) \mid a \in \{1,2,3\} \textrm{ and exactly one of} y,z \textrm{ is }6\}$, which has probability $3 \cdot \binom{2}{1} \frac{1 \cdot 1 \cdot 5}{5 \cdot 6^3} = \frac{6}{6^3}$. Here the $\binom{2}{1}$ is because the last two throws can first be a $6$, or first be something else.
The event a fair dice is $\{ (x,y,z,a ) \mid a \in \{ 1,2,3\}\}$ which has probability $\frac{3}{5}$, which gives a conditional expectation of $\frac{10}{6^3}$.