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I have met a interesting question:

If today rains, the probability that tomorrow rains is $0.6.$ If today doesn't rain, the probability that tomorrow rains is $0.2.$ Given Tuesday rained, what's the probability that Monday rained?

I have no idea how to solve this.


If I make the question a bit more complicated: Given Tuesday rain, what's the probability that the Sunday just before rained?

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    [Bayes theorem](http://en.wikipedia.org/wiki/Bayes%27_theorem) might be useful.2012-08-05
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    What is a Q? $ $2012-08-05
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    A curious FIFO...2012-08-05
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    I think you cannot tell.2012-08-05
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    I was not aware *a day* could rain.2012-08-05
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    Mainly *in Spain*.2012-08-05
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    do we consider a single week here.week starts on sunday and ends on saturday within a span of 7 days.2012-08-05

4 Answers 4

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If there is some constant probability $p$ of rain on any given day, in the absence of information about any other day's rainfall, then we have $p = 0.6p + 0.2(1-p)$, which can be solved to give $p=1/3$.

So, probability of rain on Monday and Tuesday (if we don't already know that it rained on Tuesday) is $p \cdot 0.6 = 1/5$. Probability of rain on Tuesday but not Monday (if we don't aleady know that it rained on Tuesday) is $(1-p) \cdot 0.2 = 2/15$.

But these are the cases when we KNOW that it rained on Tuesday; so the probability that there's rain on Monday given that it rained on Tuesday is $\dfrac{1/5}{1/5+2/15} = \dfrac{3}{5}$.

However, it's possible that there is no such constant probability $p$. In this case, as time goes on, the probability of rain from day to day will approach 1/3 in the limit. This is only an issue if the world has existed for a finite number of days. In this case, though, there is not enough information to solve the problem.

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    You are assuming independence...2012-08-05
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    I really don't believe that I am.2012-08-05
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    You are computing something like $p_{n+1} = 0.6 p_n + 0.2(1-p_n)$.2012-08-05
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    (Add 'I think that..." before all of my comments.)2012-08-05
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    I will admit that I am assuming that the probability of rain is equal from one day to the next - in other words, that the world isn't growing steadily drier, or steadily rainier. Is that what you mean?2012-08-05
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    Not really; the info. given is something like the probability of rain on day ${n+1}$ given that it rained on day $n$ is 0.6, as similarly for the other.2012-08-05
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    However, your assumption is a fair one, perhaps implicit in the question... I should have started by saying that you are presuming time invariance, not independence.2012-08-05
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    OK, but there'll be some asymptotic limit to the probability of rain, as time progresses. I guess I'm assuming that the equation relating probabilities from one day to the next has been true for long enough that we've reached the asymptotic limit. If you throw in the idea that the world was created last year, on a rainy day (or something similar), then this is not the case. I'll go back and edit my answer to make this assumption clear.2012-08-05
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    I'm Irish, and I know for a fact that there is no asymptotic limit to the probability of rain... :-)2012-08-05
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    @copper.hat From another (partly) Irishman - thank you for helping me to improve my answer.2012-08-05
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    +1 Thank you; it would never have occurred to me to look for a steady state. In fact, it converges to with a % of $\frac{1}{3}$ in about 5 days from any starting distribution, lending some almost Biblical credence to your assumption...2012-08-05
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    @copper.hat - Why did you remove your own answer? I found it valuable, as it came at the problem from quite a different direction from mine. It also has a certain rigour that my answer still lacks. Would you consider undeleting it?2012-08-05
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    Sure, however it is wrong in some sense...2012-08-05
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Note: The answer by David is the correct answer, it never occurred to me to look for a steady state solution to finding the probability of raining on any given day. I'm 'undeleting' this answer at David's request:

You cannot solve the problem without more information.

Let $R_M, R_T$ correspond to raining on Monday and Tuesday resp. We are given $P(R_T | R_M) = 0.6$ and $P(R_T|\overline{R_M}) = 0.2$. The desire is to compute $P(R_M|R_T) = \frac{P(R_M \cap R_T)}{P(R_T)}$.

We have $P(R_T \cap R_M) = 0.6 P(R_M)$ and $P(R_T \cap \overline{R_M}) = 0.2P (\overline{R_M}) = 0.2P (1-P(R_M))$.

Since $R_T = (R_T \cap R_M) \cup (R_T \cap \overline{R_M})$, we have $P(R_T) = 0.2 + 0.4 P(R_M)$.

Consequently, we have $P(R_M|R_T) = \frac{P(R_M \cap R_T)}{P(R_T)} = \frac{0.6 P(R_M)}{0.2 + 0.4 P(R_M)}$. Since the map $p \mapsto \frac{0.6 p}{0.2 + 0.4 p}$ is a bijection from $[0,1]$ to $[0,1]$, you can choose any $P(R_M|R_T) \in [0,1]$ and find a corresponding $P(R_M)$ that will produce your $P(R_M|R_T)$.

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Defining Events as $A$ rained on day $n$. $B$ rained on day $n+1$. We are given $P(B|A) = 0.6$ and $P(B|\bar A) = 0.2$ We have to find $P(A|B)$

Using Bayers theorem we have $$P(A|B) = \dfrac { P(B|A) \cdot P(A)} { P(B) }$$ Also $$P(B) = P(B|A) + P(B|\bar A) = 0.6+0.2 = 0.8$$ Here we don't know $P(A)$. So the problem is incomplete.

Now If one assumes $P(A) = P(B) = 0.8$ we have $$P(A|B) = \dfrac { P(B|A) \cdot 0.8} { 0.8 }$$ $$P(A|B) = P(B|A) $$ $$P(A|B) = 0.6 $$

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Let, $P(M)$ be the probability that it rains on monday, $P(T)$ be the probability that it rains on the next day which is tuesday

Let, $P(T/M)$ be the probability that it rains on tuesday given that it rained the day before, which is monday, $P(M \cap T)$ be the probability that it rains on two consecutive days, monday and tuesday

$P(T/M)=0.6$

$P(T/\overline{M})=0.2$

$P(T/M)$ and $P(T/\overline{M})$ are mutually exclusive events

i.e. $P((T/M) \cup (T/\overline{M})) = P(T/M) + P(T/\overline{M}) = 0.8$

i.e. $P(T) = 0.8$

we have from conditional probability,

$P(M \cap T) = {P(M)}{P(T/M)} = P(M)(0.6)$ -------(1)

we also have,

$P(M \cup T) = P(M) + P(T) - P(M \cap T)$ -------(2)

using equations (1) & (2), we have,

$P(M \cup T) = P(M) + 0.8 - P(M)(0.6)$

$P(M) = \frac{5}{2}P(M \cup T) - 2$ -------(3)

we, know that the probability of any event is $\le 1$

set, $P(M \cup T) \le 1$ in equation (3) to get $P(M) \le \frac{1}{2}$

i.e. $$0 \le P(M) \le \frac{1}{2}$$

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    Again, please stop using this site as your personal scratch pad. Paper is cheap. Use it.2012-08-06