Probability of choosing 5 out of 60 in ascending order.
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!
$\endgroup$ 73 Answers
$\begingroup$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.
$\endgroup$ 12 $\begingroup$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 $1/120$.
$\endgroup$ 3 $\begingroup$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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