# Probability of a random Permutation [closed]

Pick up a random permutation in S5(assuming all elements have the equal chance to be picked). Find the probability that the sum of the first three entries of σ is less than or equal to sum of last two.

My try: I mean there will be 5! different combination possible, do I have to look at each of it?

Hint: the permutations you want are just : $(...,5,4)$, $(...,5,3)$, $(...,4,5)$, $(...,3,5)$, $(...,2,5)$ and $(...,2,5)$ and you don't care about the order of the other elements where I put the dots. So the probability is: $$\frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4}.$$

• Won't we include (...,5,2) and (...,2,5) as it asks for less than or equal to?
– max
Feb 10, 2016 at 23:34
• So, there will be six cases like this ( 4 you have mentioned and 2 by me). And the probability should be (6* 3!)/120? Because we can arrange the first three elements in 3! ways.
– max
Feb 10, 2016 at 23:36
• You are right!! :D Feb 10, 2016 at 23:37
• 4+2+1 =7 ? How come is this 8?
– max
Feb 10, 2016 at 23:37
• sorry! I'm super tired (and slightly drunk). Feb 10, 2016 at 23:39