What is the probability that if $3$ of $12$ lockers are selected at random that at least two of the selections are consecutive? Three persons randomly choose a locker among 12 consecutive lockers.
What is the probability that at least two of the lockers are consecutive? 
This is what I came up with:
$|S| = C(12,3)$ all possibilities
$|E| = C(11,2) \times C(10,1) $
Thus, the probability should be 
$ P(E) = $ $C(11,2) \times C(10,1) \over C(12,3)$
Is my answer correct? 
Thanks!
 A: Your answer is incorrect.  Notice that 
$$\frac{\binom{11}{2}\binom{10}{1}}{\binom{12}{3}} = \frac{55 \cdot 10}{220} = \frac{550}{220} = \frac{5}{2} > 1$$
Method 1:  There are $12 \cdot 11 \cdot 10 = 1320$ possible ways for three people to choose their lockers.  
Now, let's count arrangements in which at least two people select consecutive lockers.  
There are $\binom{3}{2}$ ways to select two of the three people to have consecutive lockers.  There are $11$ possible starting positions for the block of two consecutive lockers and $2!$ ways to order the people within that block.  That leaves $10$ ways for the third person to select his or her locker.  Hence, there are 
$$\binom{3}{2} \cdot 11 \cdot 2! \cdot 10 = 660$$
arrangements in which at least two people have selected consecutive lockers.  
However, we have counted selections in which all three people have counted consecutive lockers twice, once when we counted the leftmost pair as having two consecutive lockers and once when we counted the rightmost pair as having two consecutive lockers.  We only want to count such selections once.  Thus, we must subtract these selections from the total.
There are $10$ possible starting positions for a block of three consecutive lockers and $3!$ orders for the people within that block.  Hence, there are 
$$10 \cdot 3! = 60$$
locker selections in which all three selected lockers are consecutive.
Hence, the number of selections in which at least two lockers are consecutive is 
$$\binom{3}{2} \cdot 11 \cdot 2! \cdot 10 - 10 \cdot 3! = 660 - 60 = 600$$
which yields a probability of 
$$\frac{600}{1320} = \frac{5}{11}$$
that at least two consecutive lockers are selected by the three people.  
Method 2:  There are $$\binom{12}{3}$$ ways to select three of the twelve lockers.  We count the number of selections in which no two lockers are consecutive, then subtract that from the total to determine the number of selections in which at least two selections are consecutive.
Line up nine blue balls in a row, leaving spaces between them and at the ends of the row in which to place three green balls. 
$$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$
That gives ten spaces in which to place the three green balls.  If we select three of those ten spaces, no two of the three green balls will be consecutive.  There are $$\binom{10}{3}$$ ways to select those three spaces.  If we now number the twelve balls from left to right, the numbers on the green balls represent a selection in which no two of the three selected lockers are consecutive.  For instance, 
$$\color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$
represents the selection of the first, sixth, and ninth lockers.  Hence, there are 
$$\binom{12}{3} - \binom{10}{3}$$
selections in which at least two consecutive lockers are selected.  Therefore, the probability that at least two consecutive lockers are selected is 
$$\frac{\binom{12}{3} - \binom{10}{3}}{\binom{12}{3}} = 1 - \frac{\binom{10}{3}}{\binom{12}{3}} = 1 - \frac{120}{220} = 1 - \frac{6}{11} = \frac{5}{11}$$
A: Suppose the individuals are A,B,C.
The number of ways of having ABC together in that order is $10!$ (similarly CAB together in that order is $10!$) so the total number of ways of having the three together is $3!\times 10!=6\times 10!$
The number of ways of having AB together in that order is  $11!$ so the number of ways of having AB together neither followed by C nor preceded by C is $11!-10!-10!=9\times 10!$ so the number of ways of having exactly two together is $3!\times 9 \times 10!= 54\times 10!$
So the number of ways of having at least two together is $54\times 10!+6\times 10!=60\times 10!$
So the probability of having at least two together is $\dfrac{60\times 10!}{12!}=\dfrac5{11}$ 
