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I am trying to understand the concept of combination and permutation. I have this situation where there are 5 men and 5 women who are ranked according to their test scores. We have the random variable X that represents the position of the higher scoring women among the ranking.

For example, P(X = 2) means that the highest scoring woman is in 2nd position. I am ask to give the answer of P(X = 6). Meaning that the highest scoring woman is in the 6th position. So other women are in 7th, 8th, 9th and 10th position respectively.

My confusion is they are using combinatorics to find a probability. They say that P(X = 6) is found by doing:

$$\frac{5!5!}{10!}$$

They explain in the textbook solution that we have 5! ways of arranging the men and 5! ways of arranging the women. Which I agree is logical. However I do not get why we divide by 10! ? 10! would represent all the possible ways of arranging the 10 people without differencing between men and women.

My confusion is that how in the world does this answer represents the fact that the highest scoring women is in the 6th positon?

My confusion is how does the answer provided before even gives a probability in the first place? We are just using combinatorics it seems. So basically we are just counting. How does it translate to a probability?

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    $\begingroup$ If all the outcomes are equally likely, then the probability of an event is equal to the number of favorable outcomes (here, $5!5!$) divided by the total number of outcomes ($10!$). This is a general principle in discrete probability. $\endgroup$ – Mike Earnest Sep 15 '19 at 17:05
  • $\begingroup$ Ah, I think you're right. I don't know how I didn't see it like that. $\endgroup$ – WindBreeze Sep 15 '19 at 18:19
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The distribution in your problem is the discrete uniform distribution. That means that there are finitely many elementary events and all of them are equally likely. Thus if $x$ is uniformly distributed on a finite set $A$ and $B \subset A$, then $P(x \in B) = \frac{|B|}{|A|}$.

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