Part one: you are planning a three-day trip to seattle, washington, in october. use the fact that on each day, it could either be sunny or rainy, and that each day is equally likely to be sunny or rainy to answer the following question. what is the probability that it is sunny all three days? part two: you are planning a three-day trip to seattle, washington, in october. use the fact that on each day, it could either be sunny or rainy, and that each day is equally likely to be sunny or rainy to answer the following question. what is the probability that it rains on at least one day?

Part one: 1/8

Part two: 7/8

Step-by-step explanation:

We can use the notatios (s,s,s) to denote that all three days are sunny, (s,r,s) to denote that the first day is sunny, the second is rainy and the third one is sunny, etc.

Using the notation above, we can describe the outcome of out trip with a sample space, that is, a set with all the possible results our trip will have.

The sample space is Ω = {(s,s,s) ; (r,s,s) ; (s,r,s) ; (s,s,r) ; (s,r,r) ; (r,s,r) ; (r,r,s) ; (r,r,r) }. Note that Ω has a total of 8 possible events that can occur and all of them with equal probability, because the probability of raining is the same than the probability of not raining. We can calculate the probability of certain event by dividing the amount favourable cases that match what the event describes with the total amount of cases from Ω, that is, 8.

PART ONE: We have only one favourable element (s,s,s). Since the probability from each element of Ω is the same, we conclude that the probability that all three days is sunny is 1/8

PART TWO: Any element of Ω that contains a rainy day (an 'r') is a favourable element. Thus, any element different from (s,s,s) is a favourable element, giving us 7 favourable cases from a total of 8. Thus, the probability is 7/8.

We could have calculated this by realizing that having al laest one rainy day is the contrary from having all three days sunny. In order to calculate the probability we can calculate instead the probability of the complementary event and substract it from 1. This is useful because we know that the probability that it is sunny all three days is 1/8, thus the probability of having at least one rainy day is 1-1/8 = 7/8.

I hope this helped you!

Step-by-step explanation:

In this case, we can put all the scenarios and then, count the scenario which can be rainy at least one day (it doesn't matter if it's day 1, day 2 or day 3 because it's asking for exactly one day). With this, we can also assume that the rest of the days will be sunny so:

We have this scenarios:

S: Sunny

R: Rainy

Now, let's do the scenarios:

If day 1 begins sunny:

SSS, SSR, SRS, SRR

If day 1 begins with Rain:

RRR, RRS, RSR, RSS

So, we have 8 possible scenarios, now let's see which ones we have one day of raining and the rest of the days, are sunny.

SSR, SRS and RSS

We have 3 scenarios, which can be rainy exactly one day.

So the probability is:

3/8 = 0.375 or simply 37.5%