I'm trying to solve some questions and I'm not sure if I'm solving it correctly.
Question 1: " A boy participating in a game and throwing normal dice, if the dice shows the number 6 the game is over. Else, the boy need to wait the number of minutes the dice shows. For example: if the dice shows 4 the boy need to wait 4 minutes and then continue playing. What is the Expected value the boy will wait?
My answer: $x$ is a random variable that gets the values ($1, \ldots, 6$) with the same probability, so the expected value should be $$2.5 = \frac{1}{6}(1+2+3+4+5) + \frac{1}{6} \cdot 0$$
Question 2: The number of earthquakes in a year at a country is a Poisson variable with $2 = \lambda$. Given that in certain year the the number of earthquakes was $2$, what is the probability that the $2$ earthquakes accord at the first $3$ months of the year?
My answer: I am not sure that my way is correct it thought that I divide $\lambda$ by $4$,then $\lambda$ is the number of earthquakes in a $1/4$ of a year and then I get that the answer is $0.0758$, but I am not sure this is the right way.
Thanks for the help!
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: $$\mathbb{E}X=15$$
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 $P\left(Y=2\mid X=2\right)$.