In rainy days, John is late for work with a probability of $0.3$. In non rainy days, he is late with a probability of $0.1$ There is a $0.7$ probability that it will rain tomorrow. Determine the probability that John won't be late for work tomorrow.
What I need is:
$$P(\lnot L|R) = \frac{P(\lnot L\cap R)}{P(R)}$$
I know $P(R) = 0.7$.
I also have
$$P(L|R)= 0.3\\ P(L|\lnot R)=0.1 \\ $$
From this I can find
$$P(L|R)=\frac{P(L\cap R)}{P(R)} \Leftrightarrow 0.3=\frac{P(L\cap R)}{0.7}\Leftrightarrow P(L\cap R) = .21$$
$$P(L|\lnot R) = \frac{P(L \cap \lnot R)}{P(\lnot R)}\Leftrightarrow0.1 = \frac{P(L \cap \lnot R)}{0.3}\Leftrightarrow P(L \cap \lnot R) = 0.03$$
So the probability of him being late should be $.21+.03=.24$. Then $P(\lnot L) = 0.76$. What is $P(\lnot L \cap R)$? I feel like I made a mistake somewhere.
Not so terribly confusing if you try thinking rather than just applying formulas!
Imagine this happening 10000 days. 70% of the time, 7000 days, it rains. The other 3000 days it does not rain. Of the 7000 days when it rains, John is late 70% of the time, 4900 days, on time the other 2100 days. Of the 3000 days when it does not rain, John is late 0.1% of the time, 3 days, not late the other 2997 days.
So of the entire 10000 days John is late 4903 days or 49.3% of the time.