Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :

\begin{aligned}k_1<k_2<k_3<k_4<k_5\end{aligned}

Also, the probability of this happening:

\begin{aligned}k_1>\max\{k_2,k_3,k_4,k_5\}\end{aligned}

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!

share|cite|improve this question
2  
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
3  
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
2  
not put back = not selected with replacement = selected without replacement. – Michael Burr Mar 27 at 19:32
up vote 6 down vote accepted

Hints:

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.

share|cite|improve this answer
    
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
1  
@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 $1/120$.

share|cite|improve this answer
    
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.

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.