# Confusion on basic probability example question

Suppose we have the following information:

1. There is a 60 percent chance that it will rain today

2. there is a 50 percent chance that it will rain tomorrow.

3. there is a 30 percent chance that it will not rain either day.

what is the probability that it will rain today or tomorrow?

Now, I don't want to retype the solution since the solution is pretty self-explanatory.

But the solution didn't really mention what is the sample space here, I was thinking the sample space is $$S=\{A, B,C\}$$ where A represents it will rain today and B: it will rain tomorrow. C: not raining on either day.

But if S is the sample space I can't really make sense between A and B, for example, I would think $$A\cap B=\emptyset$$, since raining today and tomorrow doesn't necessarily have any connection, then $$P(A\cup B)=.6+.5$$ which doesn't make any sense.

• I think it's an error to use only the events mentioned in the problem. Common sense dictates that there are four disjoint and comprehensive possibilities: it rains on neither day; it rains today but not tomorrow; it rains tomorrow but not today; and it rains on both days. If we call these $A, B, C, D \in S$ respectively, then we have $P(B \cup D) = 3/5, P(C \cup D) = 1/2, P(A) = 3/10$. Given that $P(A) + P(B) + P(C) + P(D) = 1$, it then becomes simple algebra to identify all of the probabilities. Jul 16, 2022 at 20:45
• The sample space is $\{(d,d), (d,r),(r,d),(r,r)\}$, where the first coordinate in each pair is "d" for dry, and "r" for rain today and the second, for tomorrow.
– Ruy
Jul 16, 2022 at 20:47
• Also your reasoning that "$A \cap B = \emptyset$ since raining today and tomorrow doesn't necessarily have any connection" is confused. Saying that $A \cap B = \emptyset$ is saying that $A$ and $B$ can't both happen. If $A$ is the event that it rains today and $B$ is the event that it rains tomorrow, then $A \cap B$ is the event that it rains both today and tomorrow Jul 16, 2022 at 20:48

The reason that the solution didn't mention the sample space here is that the question is a trick question.

In the present problem, you have exactly two simple events:

• $$E_1 ~: ~$$ It rains today.
• $$E_2 ~: ~$$ It rains tomorrow.

The probability of either Event $$E_1$$ occurring or Event $$E_2$$ occurring is irrelevant to solving the problem. Further, consideration of the sample space is (arguably) irrelevant to solving the problem.

Let $$F$$ denote the event that it does not rain today or tomorrow.

In effect, you are given that $$p(F) = 0.3,$$ and then asked to compute (in effect) $$1 - p(F).$$

The whole point of the problem is that the nature of the problem's $$3$$rd premise, coupled with what you are being asked to compute, renders premises 1 and 2 irrelevant.

However, in answer to your question, as the existing comments have already indicated, anytime that you are given exactly $$n$$ ($$\color{red}{\text{simple, rather than compound}}$$) events to consider (i.e. events $$E_1, E_2, \cdots, E_n$$), your sample space always has exactly $$2^n$$ elements. That is, each of events $$E_k$$ either occurs or it doesn't.

The third event discussed in the problem, that of events $$E_1$$ and $$E_2$$ both failing to occur is a compound event. That is, it is a combination of the two simple events.

Therefore, when forming the sample space, you should ignore the compound event, and enumerate your sample space as having $$2^2$$ elements, based on the simple events $$E_1, E_2$$.

"There is a 30 percent chance that it will not rain either day."

What is the probability that it will rain today or tomorrow (and possibly both)?

$$1 - 0.3 = 0.7$$