Let #d_i in {d_1,d_2,d_3}# be the number of #i# dogs a person owns where #d_3# denotes 3 or more.

A) probabilty of no dogs is #= 1 - p(d_1uud_2uud_3)#

We use the inclusion exclusion principle because we don't want to double count the homes that have 1 dog with the homes who have 2 or 3 dogs. The same is true for 2 dog owner homes.

Firs we calculate # p(d_1uud_2uud_3) = .18+.04+.01 - (.18*.04) - (.18*.01) - (.04*.01) = 13.94%#

We subtract 1 from above to find no dogs in just one of the homes. We need to have this happen in two homes so let #p(a nn b)# be the probability of event #a# and #b# where #a# is the event in the first home and #b# is the event in second. #p(a nn b) = p(a)*p(b)# if the two events are mutually exclusive and indeed knowing more about one does not help in the second. The final answer is thus #(.8606)^2 =.74063236 #

B) Some dogs would be #p(d_1uud_2uud_3)# which is just #13.94%# but again this is in one of the homes. Using the events defined above we are now interested in #p(auub)# thus #p(a)+p(b)- p(a)*p(b) = .1394+.1394 - (.1394)^2 = .25936764#

c) Dogs in each home is equivalent to finding #p(annb) = (.1394)^2 = .01943236#

d) More then 1 dog in each home would be the statement #p(annb)#. However, this time we need to calculate a different probability then the one we have been using since we are no longer interested in homes with only one dog. The update is # p(d_2uud_3)#

# p(d_2uud_3) =.04+.01 - (.04*.01) = .0496#

and #p(annb) =.0496^2 = .00256016 #