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There are two roads from $A$ to $B$, and two roads from $B$ to $C$. Each of the four roads has probability $p$ of being blocked by snow, independently of all the others. What is the probability that there is an open road from $A$ to $C$?

I got this part already. Because of independence, we have: \begin{align} P(\text{open road}) & = P(\text{open } A \text{ to } B) \cdot P(\text{open } B \text{ to } C) \\ & = (1 - p^2)^2 \end{align}

Suppose that there is a direct road from $A$ to $C$, which is independently blocked with probability $p$. What is the probability of having an open road in our five choices?

Again, because of independence, we can write: \begin{align} P(\text{all closed}) & = P(\text{orig 4 are closed}) \cdot P(\text{direct closed}) \\ & = (1 - (1-p^2)^2)p \\ & = 2p^3 - p^5 \end{align} Therefore, the probability of at least one open road is $1 - 2p^3 + p^5$. Does this look right? I found it counter-intuitive because $1 - 2p^3 + p^5 < (1 - p^2)^2$. So, the additional road decreases our chance of finding an open road?

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    $\begingroup$ Calculation looks good. Why do you say $1-2p^3+p^5<(1-p^2)^2$? $\endgroup$ – lulu Feb 26 '16 at 18:52
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    $\begingroup$ To elaborate: to see that the correct inequality goes the right way you can just ask wolfram alpha to graph them both or you can remark that $1-2p^3+p^5 = 1 -p(2p^2-p^4)> 1 - (2p^2-p^4) = (1-p^2)^2$ for $0≤p≤1$. $\endgroup$ – lulu Feb 26 '16 at 18:59
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    $\begingroup$ Keep in mind that $p \le 1.$ Thus, your inequality does not hold true. $\endgroup$ – K. Jiang Feb 26 '16 at 19:05
  • $\begingroup$ @lulu: That looks like an answer to me. $\endgroup$ – joriki Feb 26 '16 at 19:32
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As requested in the comments:

The calculation is correct. The error, such as it is, lies in the final inequality. As the OP suggests, adding an independent new road can not hurt the chances of some path from $A$ to $C$ staying open. Therefore we must have $$1-2p^3+p^5 ≥ (1-p^2)^2$$

To see that this is true we write $$1-2p^3+p^5=1-p(2p^2-p^4)$$

Now, as $p$ is a probability we have $0≤p≤1$ so $$1-p(2p^2-p^4)≥1-(2p^2-p^4)=(1-p^2)^2$$ We note further that equality can only hold here if $p=0,1$ (and indeed at these extreme cases we see that adding the extra road doesn't help).

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  • $\begingroup$ Sorry, my calculation error .... $\endgroup$ – Andy Tam Feb 29 '16 at 2:31

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