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?

• Bayes theorem might be useful. – Martin Sleziak Aug 5 '12 at 6:37
• What is a Q?  – Did Aug 5 '12 at 6:38
• A curious FIFO... – copper.hat Aug 5 '12 at 6:38
• I was not aware a day could rain. – Did Aug 5 '12 at 6:53
• Mainly in Spain. – copper.hat Aug 5 '12 at 6:58

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.

• You are assuming independence... – copper.hat Aug 5 '12 at 7:05
• I really don't believe that I am. – user22805 Aug 5 '12 at 7:05
• You are computing something like $p_{n+1} = 0.6 p_n + 0.2(1-p_n)$. – copper.hat Aug 5 '12 at 7:07
• (Add 'I think that..." before all of my comments.) – copper.hat Aug 5 '12 at 7:08
• 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? – user22805 Aug 5 '12 at 7:08

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:

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 $P(R_M|R_T)$.

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 $$P(A|B) = \dfrac { P(B|A) \cdot 0.8} { 0.8 }$$ $$P(A|B) = P(B|A)$$ $$P(A|B) = 0.6$$

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

• Again, please stop using this site as your personal scratch pad. Paper is cheap. Use it. – user5137 Aug 6 '12 at 2:10