I’m teaching a probability course and came upon a rather perplexing problem.
At an amusement park, there is a 90% chance of all rides working if it does not rain. If it rains there is a 20% chance that all rides are working. Tomorrow there is a 40% chance of rain. What is the probability that all rides are working tomorrow?
I am trying to work it as an AND or OR problem but the complexity is really throwing me off.
Does anyone know how to best approach this?
Edit: Nevermind.
In working with another math teacher, we brainstormed the following but aren’t sure how to combine them or if we even should:
(Prob Rain)(Prob Rides Work) (Prob Not Rain)(Prob Rides Work)
(.4)(.2) (.6)(.9)
Since rain and no-rain are mutually exclusive, you can just add stuff up.
Probability of rain + all rides work = 0.4 * 0.2 = 0.08.
Probability of rain + NOT all rides work = 0.4 * 0.8 = .32
Probability of NO rain + all rides work = 0.6 * .9 = .54
Probability of NO rain + NOT all rides work = 0.6 * 0.1 = .06.
Those are the four possibilities.
Then you just add the probabilities of the two results you are interested in.
Which is 0.54 + 0.08 = .62
At least that’s how I figure it.
I just thought of an easier way to do the problem.
Assume for the sake of the problem that we are looking at 100 independent days, with all the same probabilities.
In that case, it will rain on 40 days and not rain on 60 days.
Of the 40 days in which it rains, all the rides work on 8 days.
Of the 60 days in which it doesn’t rain, all the rides work on 54 days.
So out of 100 days, all the rides are working on 62.
Wow, Brazil. That makes perfect sense. It’s like you jumped inside my brain and made sense of all the static.
Thanks so much.
SDMB saves the day yet again.
This is a particular form of the law of total probability. P(A) = P(A|B)P(B) + P(A|B[sup]c[/sup])P(B[sup]c[/sup]). P(B) = .4, P(A|B) = .2 and P(A|B[sup]c[/sup]) = .9, so P(A) = .62.
I don’t mean to be rude, but this is a really basic problem. Anybody who’s teaching a probability class ought to be able to handle this.
You’re right. Normally I would be able to handle a problem like this. Like I said to Brazil, my mind is full of static right now. I’m operating on very little sleep, battling a migraine, and dealing with a new baby. My brain is complete mush. The problem makes perfect sense to me now, though.
I knew the basics of how to solve the problem, but thought that adding up the two values I found was just too simple of a solution.
Man, I posted my solution, and it was the right answer, but I edited out because I assumed it had to be more complicated and I didn’t want to look like an idiot. 
(Of course, there’s no proof of this…but I swear I got 0.62!)
That’s exactly why I thought my answer would have been incorrect, bouv.
It boils down to a simple application of the Addition and Multiplication Rules.
(Rain AND Rides Work) OR (No Rain AND Rides Work)
= (.4)(.2) + (.6)(.9) = .62 = 62%
Since any rain at all counts as rain for this problem, it is impossible for it to rain and not rain tomorrow. So the events are indeed mutually exclusive and there is no need to account for the probability of (Rain AND Rides Work) AND (No Rain AND Rides Work) which we’d normally have to consider with the Addition Rule if the scenarios were not mutually exclusive.
Without using any formulas, this problem can also be easily solved using a Venn diagram or a probability tree. For something this basic, no real need for formulas.
You are too kind.
Well, here’s your chance to earn a little extra credit:
Given that all the rides worked today, what is the probability that it rained?
There has to be something wrong with this logic, but it appears to produce the right answer. Bear in mind I have no maths training beyond high school, and I wasn’t particularly good at it even then. But here we are:
(Edit to add: please treat this as a question, not a GQ answer)
The probability of all rides working lies on a range between 20% and 90%. That is a range of 70.
The probability of rain determines where on this range the answer lies. 40% of 70 is 28.
So we reduce our probability by 40% of 70 by subtracting 28 from 90, giving 62%.
Nope, this approach actually works. It’s a little harder to understand than the “right” way, and I can’t see how to generalize to more than two states (e.g., no rain, light rain and heavy rain), but it will give you the right answer.
If anyone wants to play along at home, Paul is calculating (P(A|B[sup]c[/sup]) - P(A|B))P(B[sup]c[/sup]) + P(A|B). Showing that this is equal to P(A) is a nice little exercise in basic probability.
Let W = all rides work, R = rains. From the given info, we know P(W and R) = P(W | R)*P(R) = .20 * .40 = .08.
Since we know P(W) = .62, the answer is P(R | W ) = .08/.62 = .1290. I still prefer to use pictures, but it’s hard to post pictures.
Actually, he is calculating
P(A|Bc) - (P(A|Bc)-P(A|B))*P(B)
and this can be shown to equal P(A) (just as your expression does)
For those interested:
P(A|Bc) - (P(A|Bc) - P(A|B))*P(B)
= P(A|Bc) - P(A|Bc)*P(B) + P(A|B)*P(B)
= P(A|Bc) - P(A|Bc)*P(B) + P(A & B)
= P(A|Bc)*(1-P(B)) + P(A & B)
= P(A|Bc)*P(Bc) + P(A & B)
= P(A & Bc) + P(A & B)
= P(A)
I guess you’d have to just iterate it, starting with a pair of states and then adding others in one by one.
I used to do something like this in school to figure out what different scores on a final would give for a final grade. For example, I’d have a 75 in a class, and be writing a final worth 40% of the class grade. If I can get a 90 on the final, I’ll move 40% of the way from 75 to 90, so 81.
I kind of reasoned that answer out. That’s why I was so sure there had to be a problem - because the likelihood of a flaw in my reasoning seemed high. Of course, I hove no idea what I was actually doing in probability terms. It was just the first way that occurred to me to solve the problem.
Kinda cool that I got it right, though. I wasn’t expecting that.