Mean and Variance Calculation

HI everyone, I am having trouble with finding the answer to this question:

Suppose that 5 cities(circles) are connected by bridges (the lines in the diagram) in the following manner:


The bridge has probability p of closing down. Find the mean and variance of the number of islands that are completely cut off because of bridges closing down.
Example: So if the 4 bridges that form the x in the middle, the island in the middle (5) is cut off.


The answers are: Mean: p^3(4+p), and Variance: 4p^3(1-p^3) + p^4(1-p^4) + 8p^5(1-p^2).
I am completely puzzled and unsure where these answers came from!
Any insight would be much appreciated.

Take city 1. The probability that is is completely cut off is p x p x p = p^3 (that is all three bridges close down at the same time)

Cities 2, 3 and 4 have the same probabilities of closing down.

Whereas city 5 has a probability of p^4

Now, to set up the probability distribution function:

P(no city is closed) = (1-p)^5

P(one city is closed) = P(city one is closed or city 2 is closed or ... city 5 is closed) = p^3 + p^3 + p^3 + p^3 + p^4 = 4p^3 + p^4

P(two cities are closed) = P(1 and 2) + P(1 and 3) + ... + P(4 and 5) = p^6 + p^6 + p^6 + p^7 + p^6 + p^6 + p^7 + p^6 + p^7 + p^7 = 6p^6 + 4p^7

P(three cities are closed) = P(1, 2 and 3) + P(1, 2 and 4) + ... + P(3, 4 and 5) = p^9 + p^9 + p^10 + p^9 + p^10 + p^10 + p^9 + p^10 + p^10 + p^10 = 4p^9 + 6p^10

P(four cities are closed) = P(1, 2, 3 and 4) + P(1, 2, 3 and 5) + ... + P(2, 3, 4 and 5) = p^12 + p^13 + p^13 + p^13 + p^13 = p^12 + 4p^13

P(five cities are closed) = p^16

Mean = \(\displaystyle \Sigma xp(x) \) = 1(4p^3 + p^4) + 2(6p^6 + 4p^7) + 3(4p^9 + 6p^10) + 4(p^12 + 4p^13) + 5p^16

And I guess that the solution is considering that p^6 is so small and thus can be ignored, meaning everything larger than p^6 will also be ignored, which brings it back to
p^3 + p^4 = p^3(1+p)

Variance is the mean of the squares minus the square of the mean. The mean is p^3(1+p) and the square of the mean will be p^6(1+p)^2 = p^6(1 + 2p + p^2) = p^6 + 2p^7 + p^8.

The mean of the squares is = 1(4p^3 + p^4) + 4(6p^6 + 4p^7) + 9(4p^9 + 6p^10) + 16(p^12 + 4p^13) + 25p^16

= 4p^3 + p^4 + 24p^6 + 16p^7 + 36p^9 + 54p^10 + 16p^12 + 64p^13 + 25p^16

And the difference:

Variance = (4p^3 + p^4 + 24p^6 + 16p^7 + 36p^9 + 54p^10 + 16p^12 + 64p^13 + 25p^16) - (p^6 + 2p^7 + p^8)

= 4p^3 + p^4 - 23p^6 + 19p^7 + p^8 (disregarding anything beyond p^9)

\(\displaystyle = 4p^3\left(1 - \frac{23}{4}p^3\right) + p^4(1 + p^4) + 19p^7\)

Well, there are more assumptions that I can't seem to find, and the power 5 is what worries me. Only P(no city is cut off) gives that probability and somehow, it went there...