# Homework Help: Probability question

1. Jan 8, 2006

### gaganpreetsingh

There are n stations between two cities X and Y. At train is to stop at three of these n stations. Find the probability that no two of these three stations are consecutive.

This is what I did:

Total number of possibilities nc3 .
Now suppose that the stations train stops at is such that exactly two of these are consecutive.
(1,2) then it may stop at (4,5,6…..n) and no. of possibilities are n-3
(2,3) then it may stop at (5,6,7…..n) and no. of possibilities are n-4
…………………………………….
(n-3,n-2) then it may stop only at n and no. of possibilities are 1

So the number of ways in which the train may stop at exactly two consecutive stations is
(n-3) + (n-4) + ……… + 1 n-3 terms

using formula of A.P. [n/2 (a+l)] n is no. of terms, a is first term, l last ]
possibilities are (n-3)(n-2)/2

Now if exactly 3 stations are consecutive then (1,2,3) (2,3,4)….. (n-2,n-2,n)
Hence here the no. of possibilities are n-2

So the things I have to exclude are (n-3)(n-2)/2 + (n-2)
Which is equal to (n-1)(n-2)/2

So we have Pr{event} = 1 – (n-1)(n-2)/2*nC3 = (n-3)/n

But I am not getting the right answer. Any help where my analysis is wrong?

2. Jan 10, 2006

### EnumaElish

Let's see... Suppose the train were to make 2 stops. You could make an n-by-n table, where a row shows the 1st stop and a column shows the 2nd stop. (E.g. if n=7 then the table will have 49 cells.) Since 1st stop < 2nd stop, you need to consider the cells above the diagonal only. The consecutive stops are of the form (k, k+1); they are the cells just above the diagonal. There are n-1 such cells.

Now you could apply this logic to your case (with 3 stops).

3. Jan 12, 2006

### OptimusPrime

Gaganpreet Singh, nice to see another sardar here. WJKK WJKF! _/|\_

4. Jan 12, 2006

### EnumaElish

Do you have something to contribute to the discussion here?

5. Jan 13, 2006

### ehild

The first stop can be anything betveen 1 and n-4. Let it be "j".
The second stop "k" can be any number from j+2 to n-2.
The third stop "l" can be any number from k+2 to n.
The number of possible arrangement of stops is
$$M = \sum_{1}^{n-4} \sum_{j+2}^{n-2} \sum_{k+2}^{n}1=\frac{(n-2)(n-3)(n-4)}{6}$$

ehild