Each day is also classified as dry or rainy.The
probability of a sunny day is $0.48$, and the probability of a cloudy day is $0.39$.
The probability of having a sunny and dry day is $0.48$. The probability of a
cloudy and dry day is $0.14$. The probability of a partially cloudy and dry day
is $0.09$.
Find the probability of a cloudy day given that it's a rainy day.
My work:
Let $D$ be the event that the day is dry, $S$ be the event that the day is sunny, $R$ be the event that the day is rainy, $C$ be the event that the day is cloudy, $PC$ be the event that the day is partly cloudy.
So $P(C) = 0.39, P(S) = 0.48, P(PC) = 0.13 $
First I found Probability that it's a dry day $P(D) = P(S\cap D) + P(C \cap D) + P(PC \cap D) = .48+ .14 + .09 = .71$
Then Using the complement I found the probability that it's a rainy day $P(R) = 1 – P(D) = 1 – .71 = .29$
But How do I use this info to find $P(C | R)$?
I tried Bayes' Theorem $P(C | R) = \cfrac{P(R | C)P(C)}{P(R)} = \cfrac{P(R|C)0.39}{.29}$ But I can't find $P(R|C)$?
Can someone please tell me what I need to do to find $P(R|C)$?
Thank You!
Best Answer
It may be better to organize the data into a joint probability table. This is what we are given in the question. $$\begin{array}{c|c|c|c|c} & S = \text{Sunny} & P = \text{Partly Cloudy} & C = \text{Cloudy} & \text{Total}\\ \hline D = \text{Dry} & 0.48 & 0.09 & 0.14 & \Pr[D] \\ \hline R = \text{Rainy} & \Pr[S \cap R] & \Pr[P \cap R] & \Pr[C \cap R]& \Pr[R]\\ \hline \text{Total} & 0.48 & \Pr[P]& 0.39 & 1 \\ \end{array}$$
Since we must have the overall total equal to $1$, we require $$0.48 + \Pr[P] + 0.39 = 1,$$ or $\Pr[P] = 0.13$. We also require $$0.48 + 0.09 + 0.14 = \Pr[D],$$ so $\Pr[D] = 0.71$. This takes care of the row sums. Now for the column sums, we must have $$\Pr[D] + \Pr[R] = 1,$$ so $\Pr[R] = 1 - 0.71 = 0.29$, and similarly, $\Pr[S \cap R] = 0$, $\Pr[P \cap R] = 0.13 - 0.09 = 0.04$, and $\Pr[C \cap R] = 0.39 - 0.14 = 0.25$. So now the complete table is
$$\begin{array}{c|c|c|c|c} & S = \text{Sunny} & P = \text{Partly Cloudy} & C = \text{Cloudy} & \text{Total}\\ \hline D = \text{Dry} & 0.48 & 0.09 & 0.14 & 0.71 \\ \hline R = \text{Rainy} & 0 & 0.04 & 0.25 & 0.29\\ \hline \text{Total} & 0.48 & 0.13 & 0.39 & 1 \\ \end{array}$$ Now all that remains is to use the table to compute the desired probability: $$\Pr[C \mid R] = \frac{\Pr[C \cap R]}{\Pr[R]} = \frac{0.25}{0.29} \approx 0.862069.$$