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Suppose that whether it rains tomorrow depends on past weather conditions only through the last 2 days. Specifically, suppose that if it has rained yesterday and today, then it will rain tomorrow with probability .8; if it rained yesterday but not today, then it will rain tomorrow with probability .3; if it rained today but not yesterday, then it will rain tomorrow with probability .4; and if it has not rained either yesterday or today, then it will rain tomorrow with probability .2. Over the long term, what proportion of days does it rain

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Answer:

The answer is "0.425".

Step-by-step explanation:

The  T be a requirement today that such rains.

The Y be a situation yesterday that now it rained.

The O is a condition for tomorrow to rain.

[tex]P(O| (Y\ AND \ T)) = 0.8\\\\P(O|(Y \ AND \ NOT\ (T)) =0.3\\\\P( O|(NOT \ (Y)\ AND\ T)= 0.4\\\\P( O| NOT\ (Y) \ OR \ NOT \ (T)) =0.2[/tex]

There Arena and morrow overview To Aspects Which will rain:

[tex]Y\ AND \ T, Y \ AND \ NOT \ (T),\ NOT\ (Y) \ AND\ T, \ NOT(Y)\ OR \ NOT(T)[/tex]

Each has the option of [tex]\frac{1}{4}[/tex] INDEPENDENTLY happening.

Its possibility of tomorrow's rain is:

[tex]P(O) = P(O | Y\ AND\ T) P(Y\ AND\ T) + P(O|(Y \ AND \ NOT\ (T))[/tex]

       [tex]P((Y\ AND \ NOT\ (T)) + P(O|(\ NOT\ (Y) \ AND \ T))P(\ NOT\ (Y) \ AND\ T) \\\\+ P( O| \ NOT\ (Y)\ OR\ NOT\ (T)) P(\ NOT\ (Y)\ OR\ NOT\ (T))\\\\[/tex]

      [tex]=0.8\times \frac{1}{4} +0.3\times \frac{1}{4}+0.4\times \frac{1}{4}+0.2\times \frac{1}{4}\\\\ =\frac{1}{4}(0.8+0.3+0.4+0.2)\\\\= \frac{1.7}{4}\\\\= 0.425[/tex]