couple of questions in probability

Q1)

Let $X$ denote the waiting time and let $D$ denote the result of the first throw.

Then:

$$X\stackrel{d}{=}\left(D+Y\right)\mathsf{1}_{D\neq6}$$ where $X\stackrel{d}{=}Y$ and $Y$ and $D$ are independent.

This based on the observation that the time takes value $0$ if the first throw results in a $6$ and the process starts over otherwise in the understanding that $D$ waiting minutes are gained.

Taking expectation on both sides we find:

$$\mathbb{E}X=\mathbb{E}D\mathsf{1}_{D\neq6}+\mathbb{E}Y\mathbb{E}\mathsf{1}_{D\neq6}=\frac{5}{2}+\mathbb{E}X\cdot\frac{5}{6}$$ leading to: $$\mathbb{E}X=15$$

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Hint on Q2):

If $X$ denotes the number of earthquakes in a year then we can write: $$X=Y+Z$$ where $Y$ denotes the number of earthquakes in the first 3 months of that year and $Z$ denotes the number of earthquakes in the other months of that year.

Then $Y$ and $Z$ are independent with $Y\sim\mathsf{Poisson}\left(\frac{1}{2}\right)$ and $Z\sim\mathsf{Poisson}\left(\frac{3}{2}\right)$ .

To be found is $P\left(Y=2\mid X=2\right)$.