May 2000 15+ Challenges Solutions Rain or Shine Ling Xiang Ning from Raffles Institution, Singapore got very close to cracking this tough nut. To do so you need to use a calculator or a computer program to work out the 30 steps in the calculation for the days in June, but at each step the calculation is the same (this is called iteration). (i) p(tomorrow is dry | today is wet) = 1 - p(tomorrow is wet | today is wet) = 1- 0.7 = 0.3 p(tomorrow is dry | today is dry) = 1 - p(tomorrow is wet | today is dry) = 1- 0.2 = 0.8 (ii) Let p(June 1st is wet) = p and p(June 1st is dry) = 1- p Then p(June 2nd is wet) = 0.7p + 0.2(1- p) = 0.5p + 0.2 Also p(June 2nd is dry) = 0.3p + 0.8(1 - p) = 0.8 - 0.5p These two probabilities add up to 1. (iii) Making the simplification that each day's weather depends ONLY on the weather the day before, and using the method shown in (ii), you can find the probabilities of wet weather for each of the days of June. Whatever value you take for p on the first day, just keep multiplying by 0.5 and adding 0.2 over and over again. Let's say you know it was wet on June 1st then p=1 and you can use this in your calculation of p(July 1st is wet | June 1st is wet). For part (iv), if you knew June 1st was dry then you can use p=0 in the calculation of p(July 1st is wet | June 1st is dry) . p(June 3rd is wet ) = 0.5 (p(June 2nd is wet)) + 0.2 = 0.5 (0.5p + 0.2) + 0.2 = 0.52 p + 0.2(0.5 + 1) p(June 4th is wet) = 0.5 (p(June 3rd is wet)) + 0.2 = 0.53p + 0.2(0.52 + 0.5 + 1) What is this expression for June 30th ? , What about July 1st? For part (v) use your answers for parts (ii) and (iv). Then you can try any value of p between 0 and 1 and do the same sort of calculation for any month in the future not knowing yet what the weather will be like on the first day of that particular month. What do you discover? A:\jul00\tn386_may00.doc 2 July, 2000