# The probability of rain is 40% on Monday and 30% on Tuesday. (…) What is the probability that it will rain at least one of the two days?

My problem:

The probability of rain is 40% on Monday and 30% on Tuesday. Assume the two probabilities are independent. What is the probability that it will rain at least one of the two days?

My solution:

Saying that it will rain at least one of the two days is logically the same as saying that it'll rain on monday (event A) or it'll rain on tuesday (event B). Thus,

$$P(A \lor B) = P(A) + P(B) = 0.4 + 0.3 = 0.7$$

But...

The complement of "it will rain at least one of the two days" is that "it will not rain either day", so by the rule of complements,

$$P(\lnot A \land \lnot B) = P(\lnot A) \cdot P(\lnot B) = 0.6 \cdot 0.7 = 0.42$$

and so the probability that it will rain at least one of the two days is $$1-0.42=0.58$$

Where have I gone wrong?

• Careful with your first line of reasoning. What if I said the chance of rain on Tuesday is $70$%? By your argument, we would have $P(A)+P(B)=0.4+0.7=1.1$. Do you see the issue here? Consider using the inclusion-exclusion principle. – Theo C. Mar 24 at 23:40

The formula that $$P(A\vee B)= P(A) + P(B)$$ is only true if events $$A$$ and $$B$$ are disjoint. In this case, simply adding $$P(A)$$ and $$P(B)$$ double counts the case where both $$A$$ and $$B$$ occur.

The correct formula in this case would be $$P(A\vee B)= P(A) + P(B) - P(A\wedge B)$$, which should give you the correct answer. Note that if $$A$$ and $$B$$ are indeed disjoint, $$P(A\wedge B)$$ is $$0$$ by definition, so we can see that the first formula is just a subcase of the more general formula.

EDIT: I respond to your comment to the other answer, which suggests that it is not completely clear to you what the difference between “independent” and “disjoint” is. Independence, intuitively, means that the outcome of one event does not affect the outcome of the other. Formally, if A and B are independent, $$P(A|B)=P(A)$$ and $$P(B|A)= P(B)$$. Notably, this allows us to conclude the familiar multiplication rule for independent events: if A and B are independent, $$P(A\wedge B)= P(A)\cdot P(B)$$. To give an example, suppose that tommorow may be sunny or rainy. Also, you might win the lottery next week. These events are independent— the outcome of one will not influence the other.

On the other hand, disjoint events are events which cannot occur at the same time. In other words, if A and B are disjoint, then either A is true or B is true, (or maybe neither!), but it is not possible that both are true. Suppose that you are picking a single marble out of a bag with red and blue marbles. You might pick a red marble or you might pick a blue marble, but you can’t pick a blue marble and a red marble at the same time. Hence the events “pick a red marble” and “pick a blue marble” are disjoint.

Therefore, not only are the two ideas very different, but in fact disjoint events are never independent. This is because, for disjoint events, if A is occurs, then we know that B cannot occur. So the outcome of A can allow us to make a conclusion about the outcome of B.

Elaborating on Theo C.'s comment: the correct rule for adding probabilities is $$P(A \cap B) + P(A \cup B) = P(A) + P(B).$$ Thus, the probability of $$A$$ or $$B$$ happening is only $$P(A) + P(B)$$ if $$A$$ and $$B$$ cannot both happen at the same time. This clearly is not the case in your example.

• But the problem says that the two probabilities are independent. I see now that I wrongly assumed that A and B was disjoint and therefore $P(A \land B)=0$. But isn't that exactly what "... the two probabilities are independent" means? English is not my first language, which might be the source of my problem in this case. – Tom Mar 24 at 23:57
• Events $A$ and $B$ are independent if $P(A \wedge B) = P(A) P(B)$. – Joseph Greenwood Mar 26 at 0:34
• Your edit is completely correct. – Joseph Greenwood Mar 26 at 0:35