# Conditional Probability of rainfall

1. Apr 26, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solution
a) P(Pickwick has no umbrella | it rains) = $\frac{\frac{1}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$, which is the answer according to my answer key.

b) For part b we have:

There is a rain forecast which means he will bring the umbrella. The probability that it won't rain is 1/3.

There is a non-rain forecast which means he brings the umbrella with a probability of 1/3 and it will not rain with a prob of 2/3.

P(Pickwick has umbrella | no rain) = $\frac{1}{3} + \frac{1}{3}\frac{2}{3} = \frac{5}{9}$. But the answer is apparently 5/12. What have I done incorrectly here?

2. Apr 26, 2012

### MaxManus

I got the same as you at b) and I cant see why it is not correct.That's no guarantee for that you are correct though

3. Apr 26, 2012

### Ray Vickson

I think you are correct. Let's do it in a simple way, just by counting. Imagine 9,000 days. In 4,500 of those, it rains and in 4,500 it does not. Let's use the notation R = rain, Rc = no rain, F = Forecast rain, Fc = Forecast no rain, U = umbrella, Uc =no umbrella
Lay out the data in a table:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & N1 & N2 & (N1+N2)\\ Fc & M1 & M2 & (M1+M2) \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
We are given N1/(N1+N2) = 2/3 and M1/(M1+M2) = 1/3, as well as N1+M1 = 4500 and N2+M2 = 4500. Solving these equations we get N1 = 3000, N2 = 1500, M1 = 1500, M2 = 3000. So, the table is:
$$\begin{array}{lccc} & R & Rc & \text{tot}\\ F & 3000 & 1500 & 4500 \\ Fc & 1500 & 3000 & 4500 \\ \text{tot} & 4500 & 4500 & 9000 \end{array}$$
From this it follows that P(F) = 4500/9000 = 1/2 and P(Fc) = 1/2.

In (b), in the 4500 Rc-days, U occurs in 1500 (F-days) + (3000/3) (Fc-days), for a total of 2500 days. Thus, P(U|Rc) = 2500/4500 = 5/9.

RGV