A question about probability 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?
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.
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) &amp; (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. $$0 \le P(M) \le \frac{1}{2}$$