## 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 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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