Probability / Independence (House with Pets) The following two questions have been giving me a bit of trouble, especially (2):
1) Assume that 20% of households in general have a pet. Further assume that in a given town, a resident has three neighboring homes. What is the probability that at least one of the three neighboring homes has a pet? Assume that each neighboring home's probability of having a pet is independent of other neighboring home's probability of the same.  
2) Now assume that the independence assumption mentioned in (1) above does not hold (specifically, assume that families that have pets tend to live next to each other). How does this revised assumption affect the answer to (1)?
a) The new answer will be bigger
b) The new answer will be smaller
c) The new answer will be the same
d) Insufficient information  
For (1), I'm not too sure but this is my current thought:
$Pr(\text{at least one home w/pet}) = 1 - Pr(\text{no home has a pet}) = 1 - .80^3 = 0.488 = 48.8\%$
Please let me know if I made a mistake here.  
Now as for (2), I'm rather lost as to how to start. I suppose that if we somehow knew one neighboring home had a pet, then the other neighboring homes are more likely (> 20%) to have pets. But of course that's not the question we're being asked to solve... Another thought I had was, is the prior probability of a home's having a pet affected at all by this new assumption? I'm thinking not...then is the new answer just the same as before? Also, we're not explicitly told whether homes that do NOT have pets tend to be clustered together...is this information relevant at all here? If so I guess the answer would be (d). But I just don't know.
Would greatly appreciate any help. Thanks.
 A: For (1), you're correct. The fact that the homes are adjacent to each other isn't relevant, given that you're assuming independence.
For (2), the exact computation will depend on how you capture this correlation mathematically, but I'm tempted to write something along the lines of what you're suggesting, e.g.,
$$
 P(\text{Home has a pet} \, | \,  \text{Neighboring home has a pet}) > .2
$$
Since this is a conditional probability, the probability that a randomly selected home has a pet remains 0.2 (i.e., the prior probability remains the same), but the new assumption affects your assessment of the probability that neighboring homes have pets, once you have new information. 
If you're okay with how this dependence is quantified, then you can reason that the answer is b) as follows. The law of total probability implies
$$
 P(\text{Home has a pet} \, | \,  \text{Neighboring home } \textbf{does not} \text{ have a pet}) < .2
$$
So given that a home doesn't have a pet, neighboring homes have more than an 80% chance to not have pets either. To finish things off, note $P(\text{No home has a pet})$ can be written as a product of a prior probability and two conditional probabilities.
