Probability of a Dry Weekend
It rains on Saturday with probability $p_1 = 0.4$, independently of Sunday, which has rain probability $p_2 = 0.5$.
What is the probability that it does not rain on either day?
Hints
- Define events $R_1$ and $R_2$ for rain on each day. You want the probability that neither occurs: $P(R_1^c \cap R_2^c)$.
- If $R_1$ and $R_2$ are independent, so are their complements. Use $P(R_i^c) = 1 - p_i$.
- Multiply: $(1 - p_1)(1 - p_2) = (0.6)(0.5) = 0.3$.
Worked Solution
How to Think About It: We want both days to be dry. Since rain on Saturday and rain on Sunday are independent events, the probability of both complements occurring is just the product of each complement's probability.
Quick Estimate: Roughly half the time it does not rain Saturday (
Formal Solution:
Let $R_1$ be the event that it rains Saturday and $R_2$ the event that it rains Sunday. We want $P(R_1^c \cap R_2^c)$.
Since $R_1$ and $R_2$ are independent, their complements $R_1^c$ and $R_2^c$ are also independent. Therefore: $P(R_1^c \cap R_2^c) = P(R_1^c) \cdot P(R_2^c) = (1 - p_1)(1 - p_2)$
Plugging in $p_1 = 0.4$ and $p_2 = 0.5$: $(1 - 0.4)(1 - 0.5) = 0.6 \times 0.5 = 0.3$
Answer: The probability of a dry weekend is $0.3$.
Intuition
Independence of events carries over to their complements -- this is a basic but important fact. If knowing whether it rained Saturday gives you no information about Sunday, then knowing it did not rain Saturday also gives you no information about Sunday. Formally, $P(A^c | B^c) = P(A^c)$ whenever $A$ and $B$ are independent.
This appears constantly in risk modeling: independent failure modes, independent asset shocks, independent data sources. When you can establish independence, you can multiply probabilities -- which makes calculations tractable. The moment dependence enters (e.g., correlated rain due to a storm system spanning both days), you need the full joint distribution and the calculation becomes harder.