Statistics: Probability of two cases under specific conditions.

Question

A rain system lights up when it detects water. It performs differently under heavy and light rain. The detector fails to detect the rain 15% of the time under light rain and it fails 10% of the time under heavy rain. The light rain has a 65% chance of occurring when it rains.

My first question is what is the probability that the detector works as a whole? And the second question I have is if the detector does work, what's the probability of the rain only being light rain?

I would like to show work but this question really is not intuitive to me. I keep thinking of adding the probabilities while taking into account the fact that 65% value but I know that is not right. I'm stuck here and would really like some help. I don't even know where to start with the second question.

Thanks for any help, sorry if my question is not clear or if I should have shown some kind of work.

Answer 1

Let $D$ be the event that you detect rain. Let $L$ and $H$ be the events that you have light and heavy rain respectively.

From the problem statement we know that

• $P(D^c \mid L) = .15$ which implies $P(D \mid L) = .85$

• $P(D^c \mid H) = .1$ which implies $P(D \mid H) = .9$

• $P(L) = .65$ which implies $P(H) = .35$

Using Bayes we have

$$P(D) = P(D \mid L)P(L) + P(D \mid H)P(H)$$

$$= .85(.65) + .9(.35)$$

The second question is asking

$$P(L \mid D) = \frac{P(LD)}{P(D)} = \frac{P(D \mid L)P(L)}{P(D)} = \frac{.85(.65)}{.85(.65) + .9(.35)}$$

Let me know if you disagree! I'm learning this stuff too. =)

Answer 2

The problem could be quickly solved using a 2-way table or a tree diagram.

Here is a solution based on a 2-way table. Let's denote

• $H$ - heavy rain, $L$ - light rain
• $\color{green}{W}$ - works (detects the rain), $\color{red}{\bar W}$ - does not work (does not detect the rain)

\begin{array}{l|c|c|r} & \color{green}{W} & \color{red}{\bar W} & \\ \hline L & \color{green}{0.85 \cdot 0.65} & \color{red}{0.15 \cdot 0.95} & 0.65 \\ \hline H & \color{green}{0.9 \cdot 0.35} & \color{red}{0.1 \cdot 0.35} & 0.35 \\ \hline & \color{green}{0.8675} & \color{red}{0.1325} & 1 \end{array}

So, you get

• $P(W) = \boxed{\color{green}{0.8675}}$
• $P(L\, | \, W) = \frac{\color{green}{0.85 \cdot 0.65}}{\color{green}{0.8675}} \approx \boxed{0.637}$