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The title may be a little misleading. Let's say we choose 5 out of 60 balls. We write down the result which are in a form as $k_1,k_2,k_3,k_4,k_5$.

I have to calculate the probability of this happening :


Also, the probability of this happening:


We do care for the order so the number of the elements in the sample space is : $$\frac{60!}{(60-5)!}$$ I am stuck there. I can't think of anything to do to calculate those two probabilities. I would appreciate it if someone could help me. Thanks in advance!

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is there replacement? – Carry on Smiling Mar 27 at 19:26
I am not familiar with the english terminology, are you asking if the order matters? if that's it what you are asking then yes. – George.K Mar 27 at 19:29
The question is are the balls put back after each selection? Could $k_1=k_2$? – Michael Burr Mar 27 at 19:29
Is it possible to get the same ball twice ? – Carry on Smiling Mar 27 at 19:29
not put back = not selected with replacement = selected without replacement. – Michael Burr Mar 27 at 19:32
up vote 6 down vote accepted


Let the balls carry the numbers $1,2,\dots,60$

For the $5$ numbers on the balls that are drawn there are $5!$ orderings with equal probability and exactly one of them is an ascending order.

All $5$ drawn balls have equal chance to be labeled with the largest number.

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In the 2nd one, do we have to take the probability of one of $k_2,k_3,k_4,k_5$ being bigger that the others (to get the max) and then again the probability of $k_1$ being larger from that number? – George.K Mar 27 at 19:43
statement $k_1>\max\{k_2,k_3,k_4,k_5\}$ is exactly the same as the statement $k_1=\max\{k_1,k_2,k_3,k_4,k_5\}$. So to be found is the probability that $k_1$ is the largest. – drhab Mar 27 at 19:46
Doesn't $k_1$ have the same probability to be the largest as the other ones. But still, I can't get how we can calculate that... – George.K Mar 27 at 19:57
Yes it has. If there are $5$ candidates all having the same probability to win, then the probability that candidate1 wins is $\frac15$. This is also true for the candidates 2,3,4,5 respectively. This because you have $5$ equal probabilities that add up to $1$. – drhab Mar 27 at 20:00
@BCLC We know how the balls are drawn: at random and without replacement. Have you any reason to believe that e.g. order $5,45,7,22,31$ is more or less likely to occur then the ascending order $5,7,22,31,45$? – drhab Mar 28 at 7:02

Whatever $5$ (distinct) numbers wind up being chosen, they could have come out in any of $5!=120$ orders. So the probability they came out in increasing order is $1/120$.

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So, because all of the numbers have equal probability then the probability of $k_1 < \cdots k_5$ happening is $1/120$ ? – George.K Mar 27 at 19:39
@George.K, yes, that's right. – Barry Cipra Mar 27 at 19:43
And how about the other one. The same doesn't applie right? – George.K Mar 27 at 19:45

Each choice in which $k_1 < \cdots k_5$ corresponds one-to-one to a way to pick five balls from the set of 60. Prove this statement. Then think about how many ways there are to pick 5 balls from 60.

This should be easy. The second one is a little harder, but having thought about the first one in these terms should help. Please write back if you get stuck again.

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There are $5!$ ways to get 5 out of 60, I get that . But, that doesn't specify how many ways are in order for the first inequality to be valid. – George.K Mar 27 at 19:37
Mmm... no, there aren't. The phrase "60 choose 5" is related to this. – mathguy Mar 27 at 19:48
Wow, sorry I meant there are $5!$ ways to order the 5 taken balls. – George.K Mar 27 at 19:53

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