Using Naive Bayes to calculate the probability of user presence based on the presence of her belongings Alice, Bob, and Charlie work for a company and each of them has a cell phone and a car. They drive to work and they have their cell phones with them when they show up at work. 
The company has installed some sensors to detect the presence of the cars and cell phones. Sometimes our users use the public transportation to work or they may forget their cell phones at home. At the same time, they may leave their cell phone behind at work or use public transportation to go home. 
We interviewed the users for their habits and they say with a certain probability they are where their belongings are. I would call these probabilities P(user | belonging) which are shown in the parentheses in front of each item. 
Alice – iPhone (99%), Honda (90%)
Bob – Blackberry (98%), Ford (95%)
Charlie – Samsung (99%), Jeep (98%)
Alice is an early bird and 99% of the time, she is at work by 9:00 AM (P(Alice)=0.99). Bob usually works between 9:00 AM to 5:00 PM but he is occasionally late. We can say 80% of the time he is in the office by 9:00 AM (P(Bob)=0.8). Charlie is fashionably late and only 10 percent of the time he is in the office by 9:00 AM (P(Charlie)=0.10).
The sensors have created a database of historical information about presence of each item over time. So by 9:00 AM, the chance of each item being at the office is:
P(iPhone)=98% - P(Honda)=91%
P(Blackberry)=82% - P(Ford)=75%
P(Samsung)=9% - P(Jeep)=10%
So we want to know if Charlie’s car and cell phone are both detected by the sensors at 9:00 AM, what the probability of Charlie being at the office is.
Based on Naïve Bayes:
P(Charlie|Samsung,Jeep) = (P(Charlie) * P(Samsung|Charlie)*P(Jeep|Charlie))/(((P(Charlie) * (P(Samsung|Charlie) * P(Jeep|Charlie))+ (P(~Charlie) * (P(Samsung|~Charlie)*P(Jeep|~Charlie)))
We can calculate:
P(Samsung|Charlie) = P(Charlie|Samsung)*P(Samsung)/P(Charlie) = 0.99 * 0.09/0.1= 0.891
P(Samsung|~Charlie) = (1-P(Charlie|Samsung))*P(Samsung)/P(~Charlie) = 0.01 * 0.09/0.9 = 0.001
P(Jeep|Charlie) = P(Charlie|Jeep)*P(Jeep)/P(Charlie) = 0.98 * 0.10 /0.1 = 0.98
P(Jeep|~Charlie) = (1- P(Charlie|Jeep))*P(Jeep)/P(~Charlie) = 0.02 * 0.10/0.9 = 0.002
P(Charlie|Samsung,Jeep) = 0.1 *(0.891*0.98) /(0.1*(0.891*0.98))+(0.9*(0.001*0.002))) = 0.087318/(0.087318+0.0000018) = 0.999
Now, I have 2 questions:
1- Is this math correct?
2- If the prior probability of Jeep goes from 0.1 to 0.11(or higher), P(Jeep|Charlie) becomes 1.078(or higher). What is wrong now? 
 A: It seems you are not having a problem with Naive Bayes, but with Bayes' rule itself? Let's forget about the phones, forget about Alice and Bob. That simplifies your question to: 
given $P(C|J)=98%$, $P(J)=10%$, calculate $P(J|C)$. If you want to apply Bayes' rule here, you cannot use the value of $P(C)=10%$ that you have given, but you have to calculate it: $P(C)=P(C&J)+P(C&~J)=P(C|J)*P(J)+P(C|~J)*P(~J)$. 
However, you haven't given $P(C|~J)$. Is it supposed to be equal to $P(~C|J)$? I.e. the probability he leaves the car at work when going home is the same that he leaves the car at home when coming to work? 
A: $1$. Say we write $C$ for Charlie being at work, $S$ for his phone being there, and $J$ for his car being there. You've written $P(C|S,J) = \frac{P(C)P(S|C)P(J|C)}{P(S,J)}$. In general, it is really $P(C|S,J) = \frac{P(C)P(S|C)P(J|C,S)}{P(S,J)},$ unless you assume that $P(J|C,S)=P(J|C)$, i.e. that the chance of Charlie's belongings being at work are independent of each other. We don't have any information on how the presence of a belonging affects the chance of the other being present, so we don't have enough information to calculate $P(C|S,J)$ at all!
Still, let's see what happens if we assume they're independent. In this case I'd say your answer is roughly correct, with some inaccuracy coming from approximating $P(J|\textrm{not } C) = \frac{2}{9}$ as $0.02$. Using the exact number instead, I calculate $P(C|S,J)=\frac{87318}{87318+20} = \frac{87318}{87338}=0.999771$.
$2$. We know that $P(C,J)=P(J)P(C|J)=\frac{98}{100}P(J)$. On the other hand, $P(C)=\frac{10}{100}$. This means that our values of $P(C|J)$ and $P(C)$ put limits on how probable seeing $J$ can be, because $P(C,J)>P(C)$ would mean we are more likely to see both Charlie and his car at work than we are to see Charlie at work, regardless of where his car is. This would be absurd. What goes wrong with your example here is that the new value of $P(J)$ is higher than this imposed limit, so in practice we would simply never see this combination of three probabilities.
