I've come across a problem that I'm trying to understand:

There is a 60% chance that it will not rain on some given day, but if it does rain, then the probability of a certain amount of rainfall $R \sim Exponential(1.3)$. What is the probability that there will be less than 1" of rain tomorrow?

I approached this using conditional probability: $P(r \leq 1"|\text{Rain}) = \frac{P(r \leq 1" \cap \text{Rain})}{P(\text{Rain})} = \frac{P(r \leq 1)}{P(\text{Rain})}$.

However, computing this gave me a probability greater than 1 which doesn't make sense. Clearly the issue comes from the numerator - how do you think about the intersection between those 2 sets in a way that gives a probability of less than 0.4? Direct multiplication of probabilities should only work if $r \leq 1$ and raining are independent, which they're clearly not.

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung Oct 12 '15 at 21:41
  • $\begingroup$ think about $P(r\leq1|Rain^c)$ $\endgroup$ – Chamberlain Foncha Oct 12 '15 at 22:54

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