# Bayes theorem for calculation with personal probabilities

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?

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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.

$$Pr(L|S\cap{}E)=\frac{Pr(L\cap{}S\cap{}E)}{Pr(S\cap{}E)}$$

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.

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