Let $p \in \mathbb{N}$. Pick a number $n \in \{1, \dots, p\}$ (randomly) and toss a fair 6-sided die $n$ times. If you roll at least once the side 6, you win some money. Furthermore if you pick (at random) the number 1 and roll the sides 5 or 6, you win a bike.
(a) What is the probability of picking the number $n$?
(b) What is the probability of winning the money, given that you just picked $n$?
(c) What is the probability of winning the money?
(d) What is the probability of winning anything? (i.e. money or bike)
(e) What is the probability of having picked the number $n$, given that you just won anything?
EDIT-start
When i say "pick a number $n \in \{1, \dots, p\}$ (randomly) and toss a fair 6-sided die $n$ times", it has to be viewed as: Image an urn that contains balls with numbers $1$ to $p$ written on them. Pick one of them and assume every ball to be equaly likely to be choosen.
Lets say we pick $1$. Then we are allowed to roll a 6-sided die exactly $1$ time. If we roll $6$ we win some money and a bike. If we roll $5$ we just win the bike, but not the money. If we roll anything else, we get nothing.
Lets say we pick $3$ (and assume $p>=3$). Then we are allowed to roll a 6-sided die exactly $3$ times. If we roll at least once $6$ we win some money. But by definition we can't win the bike, because that's only possible if we had picked the ball $1$ out of the urn.
EDIT-end
I have already finished (a)-(d), but think i might have made a mistake along the road, since my answer to (e) just looks wrong.
What i've done so far:
Some notation before we start. Let $A_m$ be the event of tossing the side $m$. Let $B_n$ be the event of picking the number $n$.
(a) Since we pick the number randomly out of $p$ many, the probability is $P(B_n) = \frac{1}{p}$.
(b) It's $P(A_6 | B_n) = (1- (\frac{5}{6})^n)$, since we already know that we picked $n$.
(c) By Bayes we calculate $P(A_6 \cap B_n) = P(A_6 | B_n) \cdot P(B_n) = (1- (\frac{5}{6})^n) \cdot \frac{1}{p}$.
(d) Let $Z := A_6 \cap B_n \cup (A_6 \cup A_5) \cap B_1$ be the event of winning the money or the bike. We are looking for $P(Z) = P(A_6 \cap B_n) + P((A_6 \cup A_5) \cap B_1) - P(A_6 \cap B_n \cap (A_6 \cup A_5) \cap B_1)$. Now
... $P(A_6 \cap B_n) = (1- (\frac{5}{6})^n) \cdot \frac{1}{p}$,
... $P((A_6 \cup A_5) \cap B_1) = P(A_6 \cup A_5 | B_1) \cdot P(B_1) = \frac{2}{6} \cdot \frac{1}{p}$,
... $P(A_6 \cap B_n \cap (A_6 \cup A_5) \cap B_1) \stackrel{?}{=} P(A_6 \cap (A_6 \cup A_5) \cap B_1) \stackrel{?}{=} P(A_6 \cap B_1) = \frac{1}{6} \cdot \frac{1}{p}$.
So $P(Z) = (1- (\frac{5}{6})^n) \cdot \frac{1}{p} + \frac{2}{6} \cdot \frac{1}{p} - \frac{1}{6} \cdot \frac{1}{p} = (1- (\frac{5}{6})^n + \frac{1}{6}) \cdot \frac{1}{p}$.
(e) We are looking for $P(B_n | Z) = \frac{P(B_n \cap Z)}{P(Z)} = \frac{P(B_n \cap (A_6 \cap B_n \cup (A_6 \cup A_5) \cap B_1))}{P(Z)} = \frac{P(B_n \cap A_6 \cap B_n \cup B_n \cap (A_6 \cup A_5) \cap B_1)}{P(Z)} \stackrel{?}{=} \frac{P(A_6 \cap B_n \cup (A_6 \cup A_5) \cap B_1)}{P(Z)} \stackrel{?!}{=} \frac{P(Z)}{P(Z)} = 1$
I'm quite unsure about (d), but (e) is most likely just wrong. Where is my mistake? Thanks a lot.