Probability of experiencing rain The question is:
You are going camping over the weekend, and there is $50\%$ chance of rain on Saturday and $60\%$ on Sunday (independent). What is the probability that you will not experience rain? Then, after giving your answer, assume they are not independent, and are positively correlated. Will the chance of having a 'dry' weekend increase or decrease? 
The first part would be $0.5 \times 0.4 = 20\%$ that the weekend will not experience rain. Please help answer the second part. Thanks. 
Let's say the correlation is 0.4. Thanks. 
 A: Let $A, B$ represent the events of rain on Saturday, and on Sunday, respectively
You wish to determine the relation between the probability of a dry weekend, $\mathsf P((A\cup B)^\complement)$, and the correlation coefficient $\rho$, where the latter is:
$$\begin{align}\rho & = \mathsf {Corr}(A,B) \\ & = \frac{\mathsf P(A\cap B)-\mathsf P(A)\mathsf P(B)}{\sqrt{(\mathsf P(A)-\mathsf P(A)^2)(\mathsf P(B)-\mathsf P(B)^2)}} & = \frac{\mathsf P(A^\complement\cap B^\complement)-\mathsf P(A^\complement)\mathsf P(B^\complement)}{\sqrt{\mathsf P(A)\mathsf P(A^\complement)\mathsf P(B)\mathsf P(B^\complement)~}}
\\[1ex] & = \dfrac{100\mathsf P(A\cap B)-30}{\sqrt{6.0~}} & = \dfrac{100\mathsf P(A^\complement\cap B^\complement)-20}{\sqrt{6.0~}}
\end{align}$$


 Of course, we can get a qualitative assessment by observing that if $A$ is positively correlated with $B$, then $A^\complement$ is negatively correlated with $B$, $A$ is negatively correlated with $B^\complement$, and …  $A^\complement$ is positively correlated with $B^\complement$ .  

 Thus, if the marginal probabilities remain constant, increasing the correlation of rainy days also increases the correlation of dry days.


$\newcommand{\Chi}{\raise{0.25ex}{\chi}}$ As to the derivation of the formula, remember that: $\mathsf P(A) = \mathsf E(\Chi_A)$ so $\mathsf P(A\cap B) = \mathsf E(\Chi_A\Chi_B)$.
Then:
$$\begin{align}\rho_{A,B} & =\mathsf {Corr} (\Chi_A, \Chi_B) \\[1ex] & = \dfrac{\mathsf{Cov}(\Chi_A, \Chi_B)}{\sqrt{\mathsf{Var}(\Chi_A)~\mathsf{Var}(\Chi_B)~}}
\\[1ex] & = \dfrac{\mathsf E(\Chi_A\Chi_B)-\mathsf E(\Chi_A)~\mathsf E(\Chi_B)}{\sqrt{(\mathsf E(\Chi_A^2)-\mathsf E(\Chi_A)^2)~(\mathsf E(\Chi_B^2)-\mathsf E(\Chi_B)^2)~}}
\\[1ex] & = \dfrac{\mathsf P(A\cap B)-\mathsf P(A)~\mathsf P(B)}{\sqrt{\big(\mathsf P(A)-\mathsf P(A)^2\big)~\big(\mathsf P(B)-\mathsf P(B)^2\big)~}}
\\[1ex] & = \dfrac{\mathsf P(A\cap B)-\mathsf P(A)~\mathsf P(B)}{\sqrt{\mathsf P(A)~\mathsf P(A^\complement)~\mathsf P(B)~\mathsf P(B^\complement)~}}
\end{align}$$
Also $$\begin{align} & \qquad\mathsf P(A\cap B)-\mathsf P(A)\mathsf P(B)\\[1ex] & = \mathsf P(A)+\mathsf P(B)-\mathsf P(A\cup B)-\mathsf P(A)~\mathsf P(B) \\[1ex] & = (1-\mathsf P(A^\complement))+(1-\mathsf P(B^\complement))-1+\mathsf P(A^\complement\cap B^\complement)-(1-\mathsf P(A^\complement))~(1-\mathsf P(B^\complement)) \\[1ex] & = (1-\mathsf P(A^\complement))+(1-\mathsf P(B^\complement))-1+\mathsf P(A^\complement\cap B^\complement)-(1-\mathsf P(A^\complement)-\mathsf P(B^\complement)+\mathsf P(A^\complement)~\mathsf P(B^\complement)) \\[1ex] & = \mathsf P(A^\complement\cap B^\complement)-\mathsf P(A^\complement)~\mathsf P(B^\complement)\end{align}$$
