Probability of a certain sequence occurring from a set of sequences We have 10 towers. 7 of these 10 work while 3 do not. The system does not work if 2 or more of these not working towers appear next to each other. We need to find the probability that the given system works. Now I do know that that the total number of possible sequences are $ \binom{10}{3, 7} = 120$, but I am having trouble counting the number of sequences in which 2 or more of the out of order towers appear next to each other. Now this is what I was thinking:  


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*If we group all three bad towers together, so in 8 sequences out of 120 the system will not work.

*If we group 2 of the bad towers together. So we have $ \binom{9}{2, 7} = 36$ sequences in which the system does not work. But of these 8 were covered by the grouping of all three so:
The probability that the system works is $ \frac {120 - (36 - 8)}{120} = \frac {120 - 28}{120} = \frac {92}{120} = \frac{23}{30} $. I want to know if I am correct.

 A: As you saw, there are $\binom{10}{3}$ (that is, $120$) equally likely ways to choose where the bad towers go. Now we count the "favourables," where no two bad towers are next to each other. So we want to count the number of words of length $10$, with $7$ G's and $3$ B's, in which no two B's are next to each other. 
Line up the $7$ G's, with a little space between any two of them. They determine $8$ "gaps," (including the two endgaps). We must choose $3$ of these to slip a B into. This can be done in $\binom{8}{3}=56$ ways.  
Remark: We do it in your style. Tie two bads together, calling the result V, for very bad. Then we have a B, a V, and $7$ G's. There are $(9)(8)=72$ arrangements of these three types of letter. But each arrangement that has three consecutive B's is double counted by this $72$, so there are $72-8=64$ arrangements in which the system does not work, and therefore $56$ where the system works. 
So the approach you used does work. One disadvantage is that the approach would begin to get out of hand if we had say $5$ bad towers and $15$ good, while the "gaps" approach of the answer above generalizes smoothly.
A: Your reasoning is a bit off. 
For the system of $n$ towers, $3$ of which are bad, to be bad there is either a $bbb$ sequence ($b$ standing for bad) or no $bbb$ sequence and a $bb$ sequence.  In the former case, you have it right:  There are $n-2$ ways to locate the bad "triad".
But in the latter case, the two bads together must be followed or preceded by goods; and unless the are located at one end or another, they must be followed and preceded by goods ($gbbg$).  Counting the "at one end"
 case, you have 2 possibilities each of which has $n-3$ slots for the remaining bad tower: $2(m-3)$.  IN the non-end case, you have $n-3$ possible starting points for the $gbbg$ group, and then for each of those you have $n-4$ places to put the remaining $b$.
The total number of bad arrangements is 
$$ (n-2) + 2(n-3) + (n-3)(n-4) = (n-2)^2 $$
So for example when $n=10$ there are $64$ bad arrangements and $56$ good arrangements.
