Question

1. Consider rolling two fair, six-sided dice, and define the RV X to be the minimum of the two values you get. E.g. if s (2,5) then X(s) = min{2,5} =2. (a) (2 points) What are the possible values of X? (b) (2 points) Find the probability P(X 1) (c) (2 points) Find the probability P(X 3) (d) (4 points) Find the probability P(X 5) 2. Consider 3 events A, B, C, together with their associated indicator RV's IA, IB,Ic (a) (5 points) Express the indicator RV IAUB) as a function of the indicator RV's IA, IB (b) (5 points) Define the RV X = (IA - IB) x Ic. Express the probability P(X = 1) in terms of the probabilities of A, B, C, their unions, intersections, and complements. one can write P(IA 0) = P(Ac), or P(IA x IB 1) P(An B)) (E.g.

Answer 1

0 ----. Asa swe #------ When two dice are rolled the possible outcomes are:- 0,1,0,2), ,3), (1,4 ,1,5),(1, R 12,2), (2,3),(2,4),(2,5),(2,6) 13,1,3,2,3,3),(3,49 ,(3,5),(3,6) M,1),(4,2), (4,3), (4,49,45), (476) (511), (5,-) ,15,5), (5,4), (5,5), (576) 1611, 62, 63), (6,4),(6,5),(64) le 36 possible outcomes outcomes. Define sandom variable x = minimum of the two values (a) Minimum of (x,x) can be either ac, or 22 where both x and a2 can take values 1, 2, 3, 4, 5, 6. Therefore posible values of random variable x are 1,2,3,4156 (b) P(X=1) - favourable cases out of 36 Total no. of cases favourable cases = f (1,1%, (1,2), 0,3), (1,6, (15), 0,6), 2 (34,(3,1),(4,1),(5,12,16,1 1 = cases These fore P(x-1) = P(x=3 = favourable cases to X=3 Total no. of cases favourable cases = q (3,3),(3,4),(3,5),(3,6), ) (4,3), (513), (5,3 = 7 Cases od PCX=1 = 1 LCP 3

---- Answ4 # 2 ---- Events A, B, C are such that IA, IB IIc are their characteristic os Indicator PLIAFD) = PLASS Yoriables The indicatos variables are defined as follows: In = ş with probability Play re PLAR Iga {ó I at Ş I with probability PCB) Pege) pees Ple) Ie =şi (a) The indicator variable for AUB as defined as: I Auby = { with PLAUB) lo with PC AUB) The value of I Aus is one when IA ito and/or are having value a. and I Mursy is zero when I and IB are both zero. I puros = IA + IB X= (IA-IB) x Ic = x can take value 1 if TAXIC - I 3 I c IA XI c = 1 and If XI ( 50 R EF = PIIA=1) PCI1) RET I PLA).PLB) [ (Fox Ic) -o] - = Atleast one to zelo P(IBEO) + P(1:0) - PC Igre, Jeso) PLBY + Plce - P(B) (C) =