The probability of rain on any particular day is 0.2. However, the probability of rain on a day after a rainy day is 0.85, whereas the probability of rain on a day after a non-rainy day is 0.1 . Question in details (below)?

On three consecutive days, find the probability of having three rainy days.

2 Answers
Nov 7, 2017

It is 14.45 percent

Explanation:

The first day is independent from the day before. Therefore the first day probability (of rain) is 20 percent.

The second day (depends on the previous day) rainfall probability is 85 percent.

Finally, the third day (it must be rainy to satisfy the total probability) rainfall probability is 85 percent.

The probability of having three consecutive days which are rainy:

P(all rainy) #=0.20times0.85times0.85 = 0.1445#

or 14.45 percent.

Nov 7, 2017

The probability of having #3# rainy days is #=0.315625#

Explanation:

This is a regular Markov Chain.

The initial matrix is

#S_0=((0.2,0.8))#

The transition matrix is

#T=((0.85,0.15),(0.1,0.9))#

Therefore,

#S_1=((0.2,0.8))*((0.85,0.15),(0.1,0.9))=((0.25,0.75))#

#S_2=((0.25,0.75))*((0.85,0.15),(0.1,0.9))=((0.2875,0.7125))#

#S_3=((0.2875,0.7125))*((0.85,0.15),(0.1,0.9))=((0.315625,0.684375))#

The probability of having #3# rainy days is #=0.315625#

The infinite matrix is

#S_(oo)=((a,b))#

such that

#S_(oo)T=S_(oo)#

#((a,b))((0.85,0.15),(0.1,0.9))=((a,b))#

We have #2# equations

#0.85a+0.1b=a#, #=>#, #0.15a=0.1b#

#0.15a+0.9b=b#, #=>#, #0.15a=0.1b#

Also, #a+b=1#, #=>#, #a=2/5#

and #b=3/5#

Therefore,

#S_(oo)=((0.4,0.6))#

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