I'm completely stuck on some homework I have and can't figure it out.
The task is to calculate the probability of a bus being late conditional on the weather being snowy and bus driver being experienced.
Let $L$ be the event "The bus is late. $S$ is the event "snowy weather" and $E$ is the event "the driver is experiences.
$$Pr(L) = 0.02$$
$$Pr(L|S) = 0.4$$
$$Pr(E) = 0.6$$ (The bus driver being experienced lowers the chance of being late: personal probability)
What I have is:
$$Pr(L|S \cap{} E) = \frac{Pr(L|S)Pr(E|L\cap{}S)}{Pr(L|S)Pr(E|L\cap{}S)+Pr(\lnot E|S)Pr(E|\lnot E\cap{}S)}$$
But I'm completely stuck on setting up the calculation. How do I find $Pr(L\cap{}S)$ when I don't know $Pr(S)$, but only $Pr(L|S)$? Also, am I missing some personal probabilities? Am I even anywhere close on this?
We don't need to know the probability it is snowing, but it seems to me like we need a little more information- for instance, the probability that a driver is experienced given that it is snowing and he is late.
Start with:
$$Pr(L|S\cap{}E)=\frac{Pr(L\cap{}S\cap{}E)}{Pr(S\cap{}E)}$$
It seems reasonable to assume that that snowy weather and the experience of the driver are independent.
$$Pr(L|S\cap{}E)=\frac{Pr(L\cap{}S\cap{}E)}{Pr(S)Pr(E)}$$
We can get this into a useful form by the following steps:
$$\frac{Pr(L\cap{}S\cap{}E)}{Pr(S)Pr(E)}=\frac{Pr(L\cap{}E|S)Pr(S)}{Pr(S)Pr(E)}=\frac{Pr(L\cap{}E|S)}{Pr(E)}$$
$$=\frac{Pr(L\cap{}E|S)}{Pr(E)}=\frac{Pr(E|L\cap{}S)Pr(L|S)}{Pr(E)}$$
We know $Pr(E)$ and $Pr(L|S)$ and are just missing $Pr(E|L\cap{}S)$.
I hope this helps.