# Probability that it rains today or tomorrow

#### mthelm

##### New Member
I'm working through this nice online textbook on probability and statistics and I'm struggling to understand one of the examples:

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 does not rain either day.
Find the probability that it will rain today or tomorrow.
It seems to me that item #3 is incompatible with items #1 and #2, is it not? I thought through it like this:

The set of all possible outcomes is (today_rain, tomorrow_rain), (today_rain, tomorrow_sun), (today_sun, tomorrow_rain), (today_sun, tomorrow_sun). Given the probabilities in #1 and #2, the set looks like this: (0.6,0.5), (0.6,0.5), (0.4,0.5), (0.4,0.5). Then, we are trying to find the compliment of the probability of the event (today_sun, tomorrow_sun) which would be 1 - (0.4 * 0.5) = 0.8. However, because the author included #3, the provided answer is 1 - 0.3 = 0.7.

Can someone help me with this?

#### obh

##### Well-Known Member
Hi Mthelm,

When you calculate the 0.8 you assumed independent events!
So you just proved that the events are dependent events (it also makes sense about rain ...)
I don't really need the next following lines... but I will show the simple structure prove.

If the events are independent then P(A∩B) = p(A) * p(B) = 0.6 * 0.5 = 0.3

But P(A∪B) = 1 - P(A'∩B') by the complement rule = 1 - 0.3 = 0.7 (which is the original question's answer ...)

P(A∩B) = P(A) + P(B) - P(A∪B) by the addition rule = 0.6 + 0.5 - 0.7 = 0.4

Since P(A∩B) equals 0.4 and not 0.3 it proves that the events are dependent events.

You may also use the following calculator: https://www.statskingdom.com/probability-calculator.html

#### mthelm

##### New Member
Hi Mthelm,

When you calculate the 0.8 you assumed independent events!
So you just proved that the events are dependent events (it also makes sense about rain ...)
I don't really need the next following lines... but I will show the simple structure prove.

If the events are independent then P(A∩B) = p(A) * p(B) = 0.6 * 0.5 = 0.3

But P(A∪B) = 1 - P(A'∩B') by the complement rule = 1 - 0.3 = 0.7 (which is the original question's answer ...)

P(A∩B) = P(A) + P(B) - P(A∪B) by the addition rule = 0.6 + 0.5 - 0.7 = 0.4

Since P(A∩B) equals 0.4 and not 0.3 it proves that the events are dependent events.

You may also use the following calculator: https://www.statskingdom.com/probability-calculator.html
Oh, wow! Thank you very much for the help!!