Probability Problem

Prologue

Hi I was recently given few assignments by my master  திரு. சுதர்சன் சாந்தியப்பன் (visiting professor of SRM University), on completion of assignments he advised us to publish our works on internet. And here by I do the same with my solutions.

Assignment – 4 Probability Problem

Objective: Is to find the probability of day being rainy or dry for set of consecutive 31 day (say from 25th June to 25th July ) and then to calculate and check the probability of next five day (say 26th July to 30th July) are having following state sequence Dry , Dry, Rain, Rain

Assumption:

Considering day possess only two state namely Dry, Rain

Observation and legends:

In order to calculate the probability of day being rainy or dry for 31 days, we need to collect the actual weather conditions for these days.

Following tabular show the climatic conditions of days.

25-06-2010

Dry

26-06-2010

Rain

27-06-2010

Dry

28-06-2010

Rain

29-06-2010

Dry

30-06-2010

Dry

01-07-2010

Dry

02-07-2010

Dry

03-07-2010

Dry

04-07-2010

Dry

05-07-2010

Rain

06-07-2010

Rain

07-07-2010

Rain

08-07-2010

Rain

09-07-2010

Dry

10-07-2010

Rain

11-07-2010

Dry

12-07-2010

Rain

13-07-2010

Rain

14-07-2010

Dry

15-07-2010

Dry

16-07-2010

Rain

17-07-2010

Rain

18-07-2010

Rain

19-07-2010

Rain

20-07-2010

Rain

21-07-2010

Rain

22-07-2010

Dry

23-07-2010

Dry

24-07-2010

Dry

25-07-2010

Dry

26-07-2010

Dry

27-07-2010

Rain

28-07-2010

Rain

29-07-2010

Rain

30-07-2010

Rain

Evaluation

Let say ,

P(D) –> probability of day being Dry

P(R) –> probability of day being rainy

P (D/D) –> probabilities of day being dry preceded by the day being dry

P(R/R) –> probability of day being rainy preceded by the day being rainy

P(R/D) –> probability of day being dry preceded by the day being rainy

P(D/R) –> probability of day being rain preceded by the day being dry

Calculation:

P (D) and P(R)

Total number of days being Dry = 16

Total number of days being Rain = 15

Total number of days = 31

P(D) = 16/31 = 0.516

P(R) = 15/31 = 0.484

Transition probabilities P(D/D), P(D/R), P(R/R) and P(R/D)

Total no. of transitions = 30

No. of day being dry preceded by the day being rainy= 6

P(R/D) = 6/30 = 0.2

No. of days where climate is dry and preceding day is dry = 9

P(D/D) = 9/30 = 0.3

No. of day being rainy preceded by the day being rainy = 9

P(R/R) = 9/30 = 0.3

No. of days being rainy preceded by the day being dry = 6

P(D/R) = 6/30 = 0.2

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Probability for sequence dry dry rain rain

Based on Markov chain property

P(D,D,R,R) = P(R/R)*P(R/D)*P(D/D)*P(D)

                                      =    0.3   *   0.2    *   0.3  * 0.516

                                      =    0.009288

To test the given observations for the next 5 days

On 26th July P(D)= P (D/D) = 0.3

On 27th July P (R) = P (R/D) = 0.2

On 28th July P (R) = P (R/R) = 0.3

On 29th July P (R) = P (R/R) = 0.3

On 30th July P (R) = P (R/R) = 0.3

Result:

Thus probability of day being rainy or dry for set of consecutive 31 day (say from 25th June to 25th July ) have been calculated and probability of next five day (say 26th July to 30th July) are having following state sequence Dry , Dry, Rain, Rain have been determined

Precision =  correct events / total no .of events 

                  =  (4/5) *100 = 80%


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