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
