Probability of a particular event happening after infinite attempts I need to ask , how do u calculate the probability of a particular event will happen after infinite attempts ( at least once )
$Q.1$ Suppose I have an event with probabitlty $p$ , what is the probability that it will happen at least once .
$Q. 2$ Suppose I have an event raining .. and the probability that it rains is given by $p(n)$ for nth day .. I need to calculate the event apocalypse . apocalype happens if it rains on all days (:p) . So if I take my time period to be infinity . what is the probability that an apocalypse will happen ( wont say at least once , because if apocalypse happens .. there wont be a second time )
and of course range of $1<=n<=365$ ( no leap years )
eg. For $Q. 2$ let $p(n)=0.5$ for all days . the probability of the event apocalypse also remains $P=0.5$ instead of $P=0.5^{365}$
edit: I don't want it to rain for eternity , I just want it rain for 365 days in one calender year , once in eternity
 A: The probability that the event takes place exactly one time during an infinite sequence of independent experiments is already $1$:
$$P("\text{the number of successes is exactly one}")=p+(1-p)p+(1-p)^2p+\cdots=$$
$$=p\frac{1}{1-(1-p)}=1.$$
The probability that the event takes place at least one time is larger or equal than that but smaller or equal than $1$.
As far as raining: The daily raining events are not independent. So, we cannot use this example to enlighten ourserves.
A: Theoretically, these questions are about Tail Events, which are events defined on a sequence of random variables. The Borel-Cantelli lemmas are useful here.
Let $E_n$ be the sequence of events in Q1, then Q1 is formally asking for the probability of the tail event $$\lim\limits_{n\to \infty} \sup E_n$$
Since the $E_i$ are independent, then by the second Borel-Cantelli lemma:
$$E_i \;\textrm{independent and}\sum P(E_i)=\infty \implies P\left(\lim\limits_{n\to \infty} \sup E_n\right)=1$$
So the answer to Q1 is 1.
For Q2, let $A_i$ be the event that it rains for $365$ consecutive days starting on day $i$ of the year $i\in 1...365$. Note that all these probabilities are the same: $P(A_i)=P(A_j)=P(A)$. Now, lets define a sequence of events, $E_j\equiv A_{k},\;k=j\mod 365$. What we want to know  is 
$$P\left(\bigcap_{i=1}^{\infty} E_i^c\right)\leq P\left(\bigcap_{i=1}^{\infty} E^c_{365i}\right)=\prod_{i=1}^{\infty}P(A^c)$$
The inequality is due to the monotonicity of probability; the equality comes about because the event $E_a$ is independent of $E_{a+365}$
Therefore:
$$P(A^c)<1 \implies P\left(\bigcap_{i=1}^{\infty} E_i^c\right) =0\;\textrm{i.e., Apocalypse almost surely}$$
A: The probability that rains every day for $n$ days is $P = \left(\frac 12\right)^n$
Now if you mean "it has to rain every day of the year", then it is just $\left(\frac 12\right)^{365}$. Note that there is no "infinity" here.
If you mean "it has to rain every day for all eternity", then $n$ is not bounded by $365$ (clearly!) so the probability you're looking for is 
$$P = \lim_{n \to \infty} \left(\frac 12\right)^n = 0$$

Probability that in a given year it rains every day:
$$p = \prod_{n=1}^{365} p(n)$$
Probability that in all eternity there is at least one year when it rains every day
$$p_{ac1} = \lim_{m \to \infty} 1 - (1-p)^m$$
The probability that it rains at least one year in $m$ years is equal to $1$ minus the probability that it never rains; then take the limit.
Probability that in all eternity there is only one year when it rains every day: 
$$p_{ac2} = \lim_{m \to \infty} m p (1-p)^{m-1}$$
If $p \neq 0$ (that is to say, for all $n$ we have $p(n) \neq 0$) then you'll find $p_{ac1} = 1$ and $p_{ac2} = 0$
