The question :
Paul checks the weather report in order to decide whether to carry an umbrella. On any given day, with probability 0.2 the forecast is “rain" and with probability 0.8 the forecast is “no rain". If the forecast is “rain", the probability of actually having rain on that day is 0.8. On the other hand, if the forecast is “no rain", the probability of actually raining is 0.1.
Paul misses the morning forecast with probability 0.2 on any day in the year. If he misses the forecast, he will flip a fair coin to decide whether to carry an umbrella. (We assume that the result of the coin flip is independent from the forecast and the weather.)
Paul is carrying an umbrella and it is not raining. What is the probability that he saw the forecast?
My solution :
Let us denote by : L="Paul heard the forecast" ; U="Paul carries an Umbrella" ; R="It is raining" ; FR="the forecast is rain".
So, with these notations, I think the question is to calculate $$ P(L\,| \, U\cap R^c)$$ The Bayes formula leads to $$P(L\,| \, U\cap R^c) = \frac{P(U\cap R^c|L) P(L)}{P(U\cap R^c|L) P(L)+P(U\cap R^c|L^c) P(L^c)} $$ We have, by definition, $ P(U\cap R^c|L) P(L)=P(U\cap R^c\cap L)$ and, since if Paul heard the forecast and took an umbrella it implies that we had FR ("the forecast is rain"), this latter is equal to $P(FR\cap R^c)=P(R^c |FR)= P(R^c|FR)P(FR) = 0.2 \times 0.8=4/25$.
Now \begin{align*} P(U\cap R^c|L^c) P(L^c)=& P(U\cap R^c\cap L^c) \\ =& P(U\cap L^c )P( R^c\cap L^c) \\ = &P(U\cap L^c )P( R^c)P (L^c) \\ =& P(U | L^c )(P (L^c))^2P( R^c) \end{align*} (since $U\cap L^c $ and $ R^c\cap L^c$ are supposed to be independent and since L and R are independent (which implies $P( R^c\cap L^c)=P( R^c)P (L^c)$))
We then know that $P(U | L^c )=1/2$, since Paul flip a coin if he misses the forecast,
\begin{align*}
P(R^c)&=P(FR\cap R^c)+P(FR^c\cap R^c)\\
&= P(R^c|FR)P(FR)+ P( R^c|FR^c)P(FR^c) \\
&=0.2\times 0.2+0.8\times 0.9= 19/25
\end{align*}
and $P(L^c)=0.2=1/5$.
Gathering all that :
\begin{align*}
P(L\,| \, U\cap R^c) &= \frac{\frac4{25}}{\frac4{25}+\frac12\times\frac{19}{25}\times\frac{1}{25}} \\
& = \frac4{4+\frac{19}{50}}\\
& = \frac{200}{219}
\end{align*}
which seems to me a bit too close to 1 to be honest...
There are two scenarios in which he carries an umbrella: either he heard the forecast and it was for rain, or he skipped it and relied on the coin toss.
Scenario I: a priori, the probability he heard it is $.8$, it said rain which has a probability of $.2$, but then it failed to rain, which also has a probability of $.2$ so altogether: $$.8\times .2\times .2=\boxed{.032}$$
Scenario II: a priori, the probability that he missed the forecast is $.2$, the probability that the coin came up "bring umbrella" is $.5$ and the probability that it is not raining is given by $$.2\times .2+.8\times .9=.76$$ Thus the probability of scenario II is: $$.2\times .5\times .76=\boxed {.076}$$
By Bayes, the probability that he heard it is the portion of the total probability which is explained by Scenario I, thus: $$\frac {.032}{.076+.032}=\boxed {0.\overline {296}}$$