Sailor's weather riddle

Question

I'm stuck with this problem and right now I have no clue how to solve it. Maybe someone here might have an idea that could help solve this problem. I am not asking for a spoon-feed type of answers, I just need an explanation how, like steps on how to approach this kind of problem.

---

A ship is sailing across the ocean. The sailor is looking into the morning sky in order to get information about how the weather will turn out during the day:

Red sky at night, Sailor´s delight. Red sky in the morning, Sailor take warning.

The sailor’s weather prediction is based on observations on the morning sky.

---

Red Sky in the morning ------------------------> Storm

Not red (gray) Sky in the morning -------------> No storm

---

Assume that, on average, a storm can be expected every second day.

Red sky in the morning can be expected every fourth day and that always means a storm.

a) If the morning sky is red, how often will the sailor be right?

b) How often will the sailor be right in his weather prediction?

Answer 1

red sky $\frac{1~\text{day}}{4} $

clear sky $\frac{3~\text{days}}{4} $

storm $\frac{2~\text{days}}{4} $

Red sky & storm every $\frac{1~\text{day}}{4} $

storm no clear sky $=\frac{2~\text{days}}{4} -\frac{1~\text{day}}{4} =\frac{1~\text{day}}{4} $

$1$ | red sky and storm

$1$ | clear sky and storm

$2$ | clear sky and no storm

P(storm | clear sky) $=\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

How often with the sailor be right...

all days that are not clear sky and storm.

$\frac{3~\text{days}}{4} $

Answer 2

Let's try to make a two-by-two table. (This is often a useful approach in such problems.) It seems to me that the marginal totals are as shown. If you agree, and you can get any one of the four probabilities in the body of the table, then you can get the entire probability structure.

What is P(Red and Storm)? For which two of the four cases in the body of the table does the sailor predict correctly?

           Storm
         ----------  
Sky      Yes     No     Total
-----------------------------
Red                      1/4
Grey                     3/4
-----------------------------
Total    1/2    1/2       1 

Answer 3

$$P(R|S)=1/2=\frac{P(R\cap S)}{P(S)}$$ =>$$P(R\cap S)=1/4$$ P(S|R) implies the probability that he is right,i.e,it storms when the sky is red. So,$$P(S|R)=\frac{P(S\cap R)}{P(R)}=(1/4)/(1/4)=1$$ For the second part, he is only wrong when the sky is Grey & it storms,i.e,1/4th of the times.So he is correct 3 out of 4 times