If $\mathbb{P}(\text{Rain on Sat}) = 0.5$ and $\mathbb{P}(\text{Rain on Sun}) = 0.3$. Then if independent, $\mathbb{P}(\text{Both Days}) = \mathbb{P}(\text{Rain on Sat}) \cdot \mathbb{P}(\text{Rain on S}) = 0.15$

But if they are not independent we have $\mathbb{P}(\text{Rain on Sun | Rain on Sat }) > 0$.

What can we say for the bounds of $\mathbb{P}(\text{Both Days})$ and $\mathbb{P}(\text{Either Day})$?

  • $\begingroup$ I may not have remembered this question correclty $\endgroup$ – Permian Dec 16 '19 at 12:48
  • $\begingroup$ would a better (ie harder) question be considering prob(rain on sat OR rain on sun)? $\endgroup$ – Permian Dec 16 '19 at 16:35

Without invoking the notion of conditional probability (as in @gunes's answer) at all.....

Events $A$ and $B$ both are supersets (not necessarily proper supersets) of $A\cap B$. Consequently, $P(A\cap B) \leq P(A)$ and $P(A\cap B) \leq P(B)$ both must hold, and thus, $$P(A\cap B) \leq \min(P(A), P(B)).\tag{1}$$ No smaller upper bound is possible because the above bound $(1)$ is achievable, not just by the specific events $A$ and $B$ in the OP's question, but in general as well.

Turning to the lower bound, note that $P(A\cap B) \geq 0$ is certainly a valid lower bound for the events in the OP's question. Note, though, that it is not a bound that can only be approached arbitrarily closely as gunes says (which seems to suggest that the bound cannot be achieved exactly) but rather an exact (achievable) lower bound for the numbers in the OP's question. It is possible that it never rains on both days of the weekend and thus $A\cap B = \emptyset$ which has probability $0$. More generally, however, we have that $$1 \geq P(A\cup B) = P(A)+P(B)-P(A\cap B)$$ from which we can conclude that $$P(A\cap B) \geq P(A)+P(B)-1$$ where the right side is negative for the OP's numbers but would be positive in rainy localities where $P(A) + P(B) > 1$. We conclude that $$P(A\cap B) \geq \max(0, P(A)+P(B) -1)\tag{2}$$ and so in general, the smallest value that $P(A\cap B)$ can achieve might well be a positive number.

For the question about bounds on $P(A\cup B)$, we have the general bounds $$\max(P(A), P(B)) \leq P(A\cup B) \leq \min(1, P(A)+P(B))\tag{3}$$ which evaluate to $0.5$ (lower bound) and $0.65$ (upper bound) for the OP's numbers but both are achievable in general.

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  • $\begingroup$ thanks great answer. Im not sure how you got the RHS of (2)? $\endgroup$ – Permian Dec 17 '19 at 10:38
  • $\begingroup$ @Permian See revised answer. If you like it, you can choose to accept it by clicking the checkmark next to the answer. If a better answer comes along, you can accept the other one which will cancel the acceptance of this one. $\endgroup$ – Dilip Sarwate Dec 17 '19 at 21:09

$$\begin{align}p=\mathbb{P}(\text{Both Days})&=\mathbb{P}(\text{Rain on Sun}|\text{Rain on Sat})\times \mathbb{P}(\text{Rain on Sat})\\&=\mathbb{P}(\text{Rain on Sun|Rain on Sat})\times0.3\leq 0.3\end{align}$$

Because $\mathbb{P}(\text{Rain on Sun|Rain on Sat})$ is a probability and must be smaller than $1$. We could've done the same and obtain $p\leq 0.5$, but the former is a tighter bound.

For the minimum value, it can be arbitrarily made close to $0$.

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