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My professor mentioned that before each class, he had tea, coffee, or water. The events were set up as follows:

$P(C)=.6, P(T) = .3, P(W) = .1$ => Where C, T, and W are the probabilities of drinking Coffee, Tea or Water respectively. Now, my professor mentioned he may or may not end lecture early depending on which one he drank.

Let $l=$ event that lecture is ended early

The following probabilities are then as follows:

$P(L|C) = .2, P(L|T) = .1, P(L|W) = 0$

From this, I've managed to construct a conditional probability tree. I've also managed to figure out what $P(L \cap C) = .12$ by doing the following:

$$P(L|C)=\frac{P(L\cap C)}{P(C)}=>\frac{P(L \cap C)}{.6} =.2$$

My question now comes down to, how can I individually calculate $P(L)$ along with $P(C | L)$. I actually have no idea on how to start on these. I tried a summation of all the probabilities where L occurs, which just gives .3, but I don't feel that is correct. So to summarize my question, how can I calculate $P(L)$ and $P(C|L)$?

Using the rules for Total Probability, I've believe (without confirmation) that the following should work?

$$P(L) = P(L|C)P(C)+P(L|T)P(T)+P(L|W)P(W) => (.2)(.6)+(.1)(.3)+0=>.15$$

  • 0
    Hint: Try to find the words "Bayes formula" in your notes. – 2017-02-10
  • 0
    You can also use the "law of total probability," which allows you to compute the probability of a complicated thing by conditioning on simpler things. – 2017-02-10

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$P(L)\ne0.3$ as you suspect. The correct value may be obtained by first working out $P(L\cap T)=0.03$ and $P(L\cap W)=0$ in the same way as $P(L\cap C)$was derived. Then $P(L)$ is the sum of these three probabilities, since $C,T,W$ are mutually exclusive and no other drink is taken: $$P(L)=0.12+0.03+0=0.15$$ So your working is correct. $P(C\mid L)$ is one division away: $$P(C\mid L)=\frac{P(C\cap L)}{P(L)}=\frac{0.12}{0.15}=0.8$$