Simple question about probability I'm a bit confused about something.
Let's say I want to cross a road to my friend's house. The probability of the road to be opened is $p$, and $1-p$ to be closed. So I know the probability for one direction at a time. What would it be if I would like to go and return on the same day? 
note: Let's assume that the road was opened/closed for the whole day.  
My answer is $p^2$ (or $(1-p)^2$), depends on the status of the road.
 A: If it is given that the status of the road is the same for the entire day, then the answers of "$p^2$ or $(1-p)^2$" don't make sense. With this interpretation, only the status of the road matters, but it does not matter at all how many times you want to cross the road. With probability $p$ the road is open for the entire day — so this is the probability that you can visit your friend, or visit him and come back home, or visit him and come back home and visit him again, or … (you get the idea). And the probability that you can't cross the road is just $1-p$, regardless of how many times you want to cross it.
A: If the road is open it will be open all day and the probability that it will be open all day is $p$. And if the road is closed it will be closed all day and the probability of that is $1-p$.
So the probability you can go and return is the probability the road is open all day is $p$ and the probability you can't is the probability the road is closed all day is $1 - p$.
So I guess you question is why don't we get a paradox that the probability of GOING and  RETURNING isn't product of  Prob GOING $\times$ Prob RETURNING $= P^2$?
Well $P(A \text { and } B) = P(A)\times P(B)$ only if the events are independent of each other.  That is certainly not the case here.
To figure the probababilities of dependent events is $P(A \text{ and } B) = P(A)*(P(B|A))$  ($P(B|A)$ means the probability of $B$ given that we know $A$ happened.  In this case we know that if the road was open and went there, then the Probability that we return is $100\%$.  So $P(A\text { and } B)= p*1 = p$.
And $P(\text { not } A \text { and not} B) = P(\text {not }A)*P(\text { not }B|\text { not }B) = (1-p)*1 = 1-p$.
And by the way $P(A \text{ and not } B) = P(A)*P(\text{ not }B|P(A)) = p*0 = 0$ and $P(\text{ not } A\text { and }B) = P(\text{ not }A)*P(B|\text{ not A}) = (1-p)* 0  = 0$.
So everything works out in the end.
A: If the road doesn't change state for the whole day 
$$Pr[state=open,~on ~return|state ~on ~first ~trip]=1,$$
if $state=open$  on first trip
and zero otherwise, 
so the overall probability is $p$.
A: Let the probability the road is open be $P(O)=p$, and the probability the road is closed be $P(C)=1-p$. 
If the road is closed you can't make the journey at all and this is with $P(C)=1-p$.
If the road is open you can make the journey and this is with $P(O)=p$. Now dependent on whether the road is open or closed when you leave gives your final probability: you are stranded if the road is closed with a probaility of  $p(1-p)$, otherwise it is open and you return home with probability $p^2$. (I think this is what you meant when you said that you "know the probability for one direction at a time".)
A: I think answer is p,because who comes back and who went are the same,so he knows the road is open or closed,when the probabality of being open is P when he wants to come back anything changes,so thats just one job,so answer is P
