If it rains today, the probability that it rains tomorrow is

Question

If it rains today, the probability that it rains tomorrow is 5/10.
If it does not rain today, the probability that it rains tomorrow is 3/10.

In the long run, what is the probability that it rains on two consecutive days?

(A) 1/16
(B) 1/10
(C) 3/16
(D) 3/10
(E) 3/5

Please explain your answer.

mallelac · Answer

I am choosing E as the answer. However, my approach is very wrong I believe. On an average, half the time it may 5/10 and 3/10. Thus, I averaged these two probabilities to get 2/5 and 3/5 is close to this. hence I have chosen this. Just to let you know how I do on the exam day. Sorry for my stupid analysis 8-)

Awaiting the correct answer.

venksune · Answer

I am getting 2/5 too. 1/4 + 3/20 = 2/5. Pls see the tree diagram attached and anyone pls advise if there are any mistakes done. I am pretty sure that the logic is ok.

stuti · Answer

I am not gettin any of the ans choices above but would like to give my reasoning to know what are the holes in it..

If it is to rain on two consecutive days, say AB, there are two cases

Case1- It rains on D, A & B (D is the day before A)
Case 2- It does not rain on D, but rains on A and B

Case 1-
P(rain on A&B)=P(rains on A/rains on D)*P(rains on B/rains on A)
 =5/10*5/10 = 1/4

Case 2-
P(rain on A&B)=P(rains on A/no rain on D)*P(rains on B/rains on A)
 = 3/10 * 5/10 = 3/20

So P(rain on A&B) = 1/4+3/20 = 8/20 = 2/5

Where am I going wrong??

intr3pid · Answer

Guys, there's a very subtle trick in this question. Let's have some others contribute as well.

Venksune, to answer your question about what 's missing in that tree, simply add up all the ending probabilities. They add up to 2, or 200% probability!

venksune · Answer

No intr3pid, there is no 200% probability. Its 100% for one of the senarios ( rains  'previous day')and 100% for the other scenario (does  not rain  the 'previous day')

intr3pid · Answer

Yes, but then you're adding one probaility from the first scenario to another from the second scenario. The probability from the first scenario is already out of 100% and the same with the second one. So, when you add those probabilities, you implicitly also increase the sample space to 200%.

venksune · Answer

I agree with you Intr3pid - I have increased the sample space. Holly Molly, what am I doing...stupid me.

viengsai · Answer

I think it's 

A. 1/16

(1/8)*(5/10)

assuming all is probable, there should be 1/8 chance that it rains 2 day's straight (see Venksune's diagram). Including the condition that 5/10 of the time it rains 2 days straight, the probability becomes: 1/16

hardworker_indian · Answer

Then, I think after finding the pb, you should have divided that by 2, since the pb of raining on the"Previous day" is 1/2 (one of two possibilities - rain or no rain). I got 1/5 There are two possibilities for Previous day: a. It rains: The pb of raining in previous day = 1/2 (one of two cases - rain or no rain). The pb of raining in first day = 5/10 = 1/2 (Since it rained the Previous day). The pb of raining in second day = 5/10 = 1/2 (Since it rained the first day). Total pb = 1/8 a. It does not rain: The pb of no rain in the previous day = 1/2 (one of two cases - rain or no rain). The pb of raining in first day = 3/10 = 3/10 (Since it did not rain the Previous day). The pb of raining in second day = 5/10 = 1/2 (Since it rained the first day). Total pb = 3/40 Total pb = 1/8 + 3/40 = 1/5 (By the "Great Thoery of Choose-Maximum-Repeated", the ans is surely not 3/5 :lol:

If I do not get anywhere, I would like to choose 1/10)

intr3pid · Answer

Alright, I think it's safe to post the solution now, as, hopefully, everyone has had a chance to look at the problem.

The key to this problem is to first find out what the long-run probability of rain on any given day is.

Venksune's tree is right at the money except for the "previous day" probability which the tree leaves out. You'll notice that if you assign a probability to the "previous day" rain and no rain scenarios, the ending probabilities will all sum to 1 (i.e., 100%).

So, the paradox, then, appears to be that you need to calculate the long-run probaility which itself is needed in its calculation. If you're able to figure out that you need the long-run probability, the trick is not to let the paradox stop you but to go ahead and calculate it anyway.

So:

Long-run probability that it rains on any given day (say, today) =

(Given that it rained yesterday in the long run, it rains today) + (Given that it did not rain yesterday in the long run, it rains today)

Let p = long-run probability of rain.

Then, based on the above equation, we have:

p = (p)(5/10) + (1-p)(3/10) = (5p + 3 - 3p)/10 = (2p + 3)/10
=> 10p = 2p + 3
=> p = 3/8

i.e., Probability that it rains on any given day in the long run = 3/8

Then, probability that it rains on two consecutive days in the long run =

Given that it rains on a given day in the long run, it rains the next day =

(3/8)(5/10) = 3/16. Which is C.

Notes:

For those of you who don't have maths/stats backgrounds, the long-run probability of any event is called its limiting probability. You use a technique called Markov Chains (in the form of matrices) to calculate limiting probabilities. The problem above is an over-simplified version of a Markov Chain, which you may run into at the b-school. If you do, feel free to return to this problem and re-solve it.

This question appeared on a preliminary actuarial exam in 2000, so you shouldn't sweat it if you couldn't solve it. I've only posted here so that you guys can get practice with some unusual probability questions.

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This question appeared on a preliminary actuarial exam in 2000, so you shouldn't sweat it if you couldn't solve it. I've only posted here so that you guys can get practice with some unusual probability questions.

If it rains today, the probability that it rains tomorrow is

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