My Math Forum Conditional Probability - Rainy days

 November 11th, 2014, 02:14 PM #1 Newbie   Joined: Nov 2014 From: Chicago Posts: 2 Thanks: 0 Conditional Probability - Rainy days The Question: The weather in Kuala Lumpur behaves as follows: • Each day is either warm and sunny or warm and rainy, • the first day of each year is rainy or sunny with probability 1/2 each, • every other day is the same kind as the day before with probability 2/3 and the other kind with probability 1/3. For a year containing 365 days, define the random variables Xi to be 1 if the weather is sunny on the i-th day and −1 otherwise. Meaning Xi = 1 if day i is sunny -1 day i is rainy Also, let Yi = { X1 if i=1 Xi· X(i−1) else } a. Write the number of rainy days throughout the year as an expression over the Xi variables. b. Compute the expectation of the number of rainy days throughout the year. c. Show that the variance of the number of rainy days in the year is bounded by 150. d. Use the variance to compute an upper bound on the probability that the number of rainy days throughout the year exceeds 250. My Answer is as follows and i want to know if it is correct: a. The number of rainy days is 365 - the sum from 1-365 of ((i+1)/2) what i did was turn the values of Xi to 1 if sunny and 0 if rainy. the sum gives me the total of sunny days, therefore the rest are rainy. b. P(X1 = 1) = 1/2 P(X2 = 1) = 2/3(P(X1 = 1) + 1/3(P(X1 = -1) this equals 2/3*1/2 + 1/3*1/2 which equals 1/2. P(X3 = 1) = 2/3(P(X2 = 1) + 1/3(P(X2 = -1) this equals 2/3*1/2 + 1/3*1/2 which equals 1/2. Basically, if we use Inductive reasoning, then p(Xn = 1) = 1/2. Therefore, I can define this problem as a Binomial model problem, which is why E(RainyDays) = NP = 365*1/2 = 182.5. C. Var(RainyDays) = NP*(1-P) = 182.5/(1/2) = 91.25 we need to prove Var(RainyDays) < 150 which it is. D. P(Number of Rainy Days >250) = ? Chebyshev says: P(|X-E(X)| >= k) <= Var(X)/(K^2) so we can use k = 67.5 P(|X - 182.5| >= 67.5) <= 91.25 / (67.5^2) P(X>= 250) <= 91.25/4556.25 = 0.02. Now, This seems to me like its not the answer they are looking for, but is it correct? what other way is there to do this? Thanks!!
 November 29th, 2014, 12:09 AM #2 Senior Member   Joined: Aug 2012 Posts: 229 Thanks: 3 Hey NRS. Are you still interesting in a response? If so please reply.

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