How many days should i pay for staying at the hotel before my trip? I live in city $B$. Assume that i'm going on a vacation to city $A$. Before i go there, for every night i should pay 60 dollars and after i go there, for every other night i should pay 100 dollars.    
After i go there, the company i work for, might call me for something important to return to city $B$. The probability that this happens in $0.3$ for every day.  
Question : How many days it's better for me to pay for ? (before going to the trip)
Note : I have no idea about this question and i don't know how that probability is related to the question.
Thanks in advance.
 A: The chance your boss calls specifically on day $n$ is $0.7^{n-1}\cdot 0.3$, where the factor $0.7^{n-1}$ comes from the fact that he must not have called before.  As the comments say, this is a geometric distribution.  You should clearly prepay the first night, as you are guaranteed to need that hotel room.  If you prepay the second night, after two nights you will have spent $120$.  If you only prepay the first night, you have $0.3$ chance of only staying one night and spending $60$ and $0.7$ chance of staying at least two nights and spending $160$ in the first two nights.  This gives an expected cost of $0.3\cdot 60+0.7\cdot 160=130$, so you are ahead paying for two nights.  You can do a similar calculation for three nights.  I am sure it will be a losing proposition to pay for three nights.
A: My understanding of the question is that you plan to stay on vacation until your company calls you back, and you want to minimise the expected cost.
On the assumption that the recall events are independent, the probability that you'll be staying after $k$ recall opportunities (which, if I understand correctly, occur before the $k$-th night) is $p_k=\left(\frac7{10}\right)^k$. If you pay in advance, the cost for this night is $\$60$ independent of whether you stay (assuming that you can't get a refund, which would render the question pointless). If you don't pay in advance, the expected cost is $\$100\cdot p_k$. Thus you should pay in advance if $p_k\gt\frac35$, which is the case for $k\le1$. Thus you should pay in advance for all nights up to the second recall opportunity.
A: You will definitely need to one night in the hotel,  and you have a 70% chance that you will stay a second night.  But you only have a $(0.70)^2 = 0.49\%$ chance of needing a 3rd night.
The exected cost of an additional night is $\$60$ if you prepay and $(0.70)^{n-1}\cdot \$100$ if you pay as you go.
When $(0.70)^{n-1} > 0.60$ pre-pay (second night night)
When $(0.70)^{n-1} < 0.60$, pay as you go (3rd and beyond). 
