Assume there are 3 balls in a box, two of them have the digit $1$ printed and the other one has digit $2$ printed.
Let $ X $ be a random variable that denotes which ball was chosen.(uniform probability)
If $X=1$ we roll a dice until we get an even result.
Else we roll a dice until we get 6. Let $ Y $ be the amount of dice rolls required.
Calculate $$ \mathbb{P}(Y\geq3|X=1) $$
Basically we need $$ \frac{\mathbb{P}\left(Y\ge3\cap X=1\right)}{\mathbb{P}\left(X=1\right)} $$ How do I solve this using Bayes law, or how can I apply the $\mathbb{P}(A)=1-\mathbb{P}(A^C)$?
What I wanted to do is $$ \frac{1-\mathbb{P}\left(Y=2\cap X=1\right)-\mathbb{P}\left(Y=1\cap X=1\right)}{\mathbb{P}\left(X=1\right)} $$ This is wrong because $$ \left(Y\ge3\cap X=1\right)^{C}=\left(\left(Y=anything\cap X=2\right)\cup\left(Y=1\cap X=1\right)\cup\left(Y=2\cap X=1\right)\right) $$ But then I would need to work hard for $X=2$ so I figured that the whole $X=1$ experiment has probability $\frac{2}{3} $ So I can just calculate $$ \frac{\mathbb{P}\left(Y\ge3\cap X=1\right)}{\mathbb{P}\left(X=1\right)}=\frac{\mathbb{P}\left(X=1\right)-\mathbb{P}\left(Y=2\cap X=1\right)-\mathbb{P}\left(Y=1\cap X=1\right)}{\mathbb{P}\left(X=1\right)} $$ Why does it work out like that?Is there a way to solve this with known formulas?
Let $Z_1, Z_2, Z_3, \ldots$ be the sequence of rolls (i.e. $Z_i$ is the number of the $i$th roll). Then
$$P(Y \geq 3 \mid X=1) = P(Z_1 \text{ and } Z_2 \text{ are both odd } \mid X=1),$$
since all the $Z_i$ are independent of $X$, $$P(Z_1 \text{ and } Z_2 \text{ are both odd } \mid X=1) = P(Z_1 \text{ and } Z_2 \text{ are both odd}),$$
and since $Z_1, Z_2$ are independent, $$P(Z_1 \text{ and } Z_2 \text{ are both odd}) = P(Z_1 \text{ is odd})\cdot P(Z_2 \text{ is odd}).$$
That is, $$P(Y \geq 3 \mid X=1) = P(Z_1\text{ is odd})\cdot P(Z_2\text{ is odd}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$