Probability of A given independent variables Sorry I am a total novice in statistics and cannot find my answer anywhere.
I want to know if there is a way of calculating the probability of a binary outcome given a number of independent variables.
Example:
Imagine I am interested in predicting the probability that someone will be late to work.
I have data on the effect the following three variables have on the likelihood someone is late:

*

*The probability of being late given someone has children is 0.3

*Probability of being late given someone takes the bus is 0.5

*Probability of being late given their alarm has a snooze function is 0.1

I do no have any data about the interdependence of these variables.
Is it valid to compound these probabilities in some way to give an indication?
Say someone takes the bus and has a snooze function, what would be the probability of them being late given the probabilities given above?
 A: An set that is exhaustive (covers the entire sample space), and mutually exclusive (its members do not overlap) is called a partition. One partition of the sample space is {unemployed, employed hourly, employed salaried}, assuming there are no other types of employment, so everyone is either unemployed, hourly, or salaried. We can then use the Law of Total Probability to state $$P(late)=P(late \cap unemployed)+P(late \cap  hourly)+P(late \cap  salaried)=P(late|unemployed)*P(unemployed)+P(late|hourly)*P(hourly)+P(late|salaried)*P(salaried)$$
Now back to your example. Because {has children, takes the bus, has alarm with snooze function} is not a valid partition, because the outcomes are not mutually exclusive, you cannot say:
$$P(late)\not=P(late \cap has children)+P(late \cap takes the bus)+P(late \cap has alarm with snooze function)$$
You ask. "Say someone takes the bus and has a snooze function, what would be the probability of them being late given the probabilities given above?" There is not enough information to answer this question because we do not know how the event "takes the bus" is related to the event "has a snooze function".
