Probability Question I can't get around. This is the question from my assignment, which I can't get around. 
Suppose that a water distribution system is composed of a number of independent 
pipes. At temperatures below 0 deg C, the pipes may burst; on such occasions the 
probability that one pipe will catastrophically burst is 0.05. Assume that the failure 
(catastrophic bursting) of two or more pipes at the same time is unlikely. If the 
distribution system must be shut down when 3 of the pipes have failed, determine the 
probability that the distribution system can withstand at least 5 below 0 deg C events 
before a shut down occurs.
How do you do this question without knowing number of pipes ? I understand how it is possible if one knows the number of pipe. But how do you do it if number of pipes are not given ?
 A: So there are $2^5$ possible sequences in terms of break or not break.  16 of those 32 have 3 or more breaks.  More precisely, there are 10 ways to get 3 breaks, 5 ways to get 4 breaks, and 1 way to get 5 breaks.  However, these sequences are weighted.  The easiest is the 1 way to get 5 breaks, this probability is 0.05^5.  The second easiest is 5 ways to get 4 breaks, this can be any of
$$\begin{matrix}
oBBBB \\
BoBBB \\
BBoBB \\
BBBoB \\
BBBBo  \\
\end{matrix}$$
where $B$ is a break and $o$ represents no break.  The probability of each of these is
$$
\begin{matrix}
  0.95 * 0.05* 0.05* 0.05* 0.05 = 5.9375 * 10^{-6} \\
  0.05 * 0.95* 0.05* 0.05* 0.05 = 5.9375 * 10^{-6}\\
  0.05 * 0.05* 0.95* 0.05* 0.05 = 5.9375 * 10^{-6}\\
  0.05 * 0.05* 0.05* 0.95* 0.05 = 5.9375 * 10^{-6}\\
  0.05 * 0.05* 0.05* 0.05* 0.95 = 5.9375 * 10^{-6}\\
\end{matrix}
$$
You have 10 ways to obtain 3 breaks out of 5, and perhaps there is a pattern with those as well ;)  
The probability of having NO break will be 1 minus the sum of all the breaking scenarios above (I get 99.88% chance you can withstand going below zero 5 times).  
