There is of course, a brute force way to answer this problem if, as @glugglug asked, the probability of a bridge breaking is not 0.5.
First, what's the probability of the three eastmost bridges all still standing? This gives us a path from North to South using only three bridges. Well, if the probability of a bridge still standing is p (in our case, p=0.5), the probability of all three of those bridges standing is p^3 (in this specific case, 0.125).
If those bridges are up, great, we will take those, but in the remaining 0.875 cases (1-p^3), what are the chances that the three center bridges are standing? Again, that's p^3, but it's p^3 out of the remaining 1-p^3 cases, not out of 1, because we wouldn't be looking at the center bridges if we had already found a way across on the eastmost ones. So, the chances of either being able to cross the eastmost briges or the center bridges are (p^3) + (p^3) * (1 - p^3).
Or, more generally, (probability of all the previous routes) + (probability of current route)*(1 - probability of all the previous routes).
For the westmost bridges, we would do the same thing, the probability of those three bridges standing is p^3, the total probability of a route using only three bridges existing is
[(p^3) + (p^3) * (1 - p^3)] + (p^3) * [(p^3) + (p^3)*(1 - p^3)].
You have to continue summing these up until you have accounted for all possible routes, I believe there are 3 that only use 3 bridges (each with probability p^3), 8 that use 4 bridges (p^4), 4 that use 5 bridges (p^5), 4 that use 6 bridges (p^6), and 4 that use 7 bridges (p^7), but I just did that very quickly in my head and may have missed some.
If anyone wants to do all that out by hand with p=0.5 and see if the answer is also 0.5, be my guest, I may do it later if I'm bored enough.