Conditional Probability of rainfall

[TranscendArcu]

Homework Statement

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?

 

[MaxManus]

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

 

[Ray Vickson]

Homework Statement

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?

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}<br /> &amp; R &amp; Rc &amp; \text{tot}\\<br /> F &amp; N1 &amp; N2 &amp; (N1+N2)\\<br /> Fc &amp; M1 &amp; M2 &amp; (M1+M2) \\<br /> \text{tot} &amp; 4500 &amp; 4500 &amp; 9000<br /> \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}<br /> &amp; R &amp; Rc &amp; \text{tot}\\<br /> F &amp; 3000 &amp; 1500 &amp; 4500 \\<br /> Fc &amp; 1500 &amp; 3000 &amp; 4500 \\<br /> \text{tot} &amp; 4500 &amp; 4500 &amp; 9000<br /> \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