Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 25% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let X denote the number among the four who have earthquake insurance. a. ?Find the probability distribution of X. [Hint: Let S denote a homeowner who has insurance and F one who does not. Then one possible outcome is SFSS, with probability (.25)(.75)(.25)(.25) and associated X value 3. There are15 other outcomes.] b.? ?Draw the corresponding probability histogram. c.? ?What is the most likely value for X? d.? ?What is the probability that at least two of the four selected have earthquake insurance?

Solution : Step 1 : Here they randomly selected one metropolitan area in california, where 25% of all homeowners are insured against earthquake. From that four homeowners are to be selected at random, and X denote the number among four who have earthquake insurance. Step 2: a) We have find the probability distribution of X. it is clear that for this event we have two possible outcomes.when we selected homeowner either they are insured or not. We will let S denote a homeowner who has insured and F one who does not. It is given that P(S)= .25 P(F) = .75 It is also given that there is 16 possible outcomes for this event. So by taking 4 homeowners at random. We can list all possible outcome with their probabilities. So P(x=0) = P(FFFF) this is the probabilities that none of the homeowners are insured. 4 = (0.75) = .31640 P( X=1) = P(SFFF)+P(FSFF)+P(FFSF)+P(FFFS), this is the probability that one of the homeowners insured others not. There is 4 possible ways for this so, = 4× (0.25)×(0.75) 3 = 0.421 P(X=2) = P(FFSS)+P(SSFF)+P(FSSF)+P(SFFS)+P(FSFS)+P(SFSF) , this is the probability that 2 of them are insured 2 of them not. There is 6 possible outcomes for this , so = 6× (0.25) × (0.75) 2 = .2109 P(X=3) = P(SSSF)+P(FSSS)+P(SFSS)+P(SSFS), this is the probability that 3 of them is insured one is not, there is 4 possible outcomes for this, so 3 = 4× (.25) × (0.75) = .04687 P( X= 4) = P(SSSS) , this is the probability that all of the selected homeowners are insured. = (0.25)4 = .003906 By tabulating this probability distribution we will get X 0 1 2 3 4 P(X) 0.31640 0.421 0.2109 .04687 .003906 By adding the probability we will get one.