# Thread: Conditinal Probability with numbers selected at random

1. ## Conditinal Probability with numbers selected at random

Suppose that three numbers are selected one by one, at random, and without replacement from the set of numbers {1, 2, 3, ... n}. What is the probability that the third number falls between the first two if the first is smaller than the second?

The book tells me that the answer is 1/3, but that isn't much help towards how to find the answer.

2. Hello, Zennie!

$\displaystyle \text{Suppose that three numbers are selected randomly without replacement}$
$\displaystyle \text{from the set of numbers }\{1,\,2,\,3,\,\hdots\, n\}.$

$\displaystyle \text{What is the probability that the third number falls between the first two}$
$\displaystyle \text{if the first is smaller than the second?}$

$\displaystyle \text{The book answer is: }\frac{1}{3}$

Let the three numbers be: .$\displaystyle a,\,b,\,c$

There are $\displaystyle 3! = 6$ permutations of the three numbers:
. . $\displaystyle abc,\:acb,\:bac,\:bca,\:cab,\:cba$ . (in increasing order)

"The first is less than the second": .$\displaystyle a < b$
. . There are three cases: .$\displaystyle abc,\:acb,\:cab$

In only one case, $\displaystyle \,c$ is between $\displaystyle \,a$ and $\displaystyle \,b$: .$\displaystyle acb$

Therefore, the probability is: .$\displaystyle \dfrac{1}{3}$

3. Wow, thanks a lot. That's so much more simple than I was expecting. I probably wouldn't have thought about it that way.

4. It is not such a tough question. Try using this hint:
Pick three number (in order) - x1,x2,x3 from {1,2,3,.....,n}
What is the Prob(x1<x2<x3)? [You don't need any calculation - think symmetry !!]