Clarification on Probability distributions

if possible id like it if people could just clear up a couple of issues i'm having but first context!

i'm currently studying Probability from "a first course in probability" chapter 4 - Random Variables.

throughout the chapter i've learnt how to calculate the expectation when given a random variable $X$ as well as the variance. i've had a run through the bernoulli and binomial random variables (including properties), the poisson random variable, the geometric random variable, the negative binomial random variable. the hypegeometric random variable, the negative bionomial random variable as well as calculating the expectation of a sum of random variables. (all of which have been single variable)

now im onto the chapter problems and thus this leads onto my issue:

Question 4.4: Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman. (For instance, X = 1 if the top-ranked person is female.) Find $P\{X = i\}$, $i = 1, 2, 3, . . ., 8, 9, 10$.

My assumption is that i need to use one of the random distributions covered by the chapter but i'm completely unable to, i have no idea how to proceed in this direction.

i understand that the scores of the individuals are distinguishable, that any 1 combination has a probability of $\frac{1}{10!}$ chance of occuring and that this will have some sort of effect on X our random variable. in order to solve this problem i went about it this way:

for the probability of a women being in the top spot we have 2 groups of which they have equal probability so this leaves us with a cool nice $\frac{1}{2}$

a women being in the second spot we think of it as, a man must have taken the position 1 then any of our 5 women left over and we cycle through them.

this is equivilant to letting $E_{i}$ be the event that the women holds the i'th spot. then $$P(E_{2}|E_{1}^c)=\frac{P(E_{2}\cap E_{1}^{c})}{P(E_{1}^{c})}$$ and so on leaving us with $$P\left(E_{i}|\bigcap_{n=1}^{i-1}E_{n}^{c}\right) = ~ ....~$$ etc etc etc.

but again this method doesnt use the idea of a random variable nor any of the distributions.

this led me to considering how i would generate a function (whilst ignoring the distributions that i know) for the $X=i$ (ie using the random variable)

this led me to

$$P(X=i) = \frac{{{5}\choose{1}}{{5}\choose{i-1}}(i-1)!(10-i)!}{10!}$$

which is the number of ways of choosing 1 women to fill the ith spot, the number of i-1 men infront of her, multiplied by the number of ways of re-organising said men and finally multiplying by the number of combinations of (10-i) people after our top scoring women all over the sample space...

so this does work (at least according to the answers) my issue is this isn't one of the many distributions which was covered throughout the chapter. is there a way of modeling this problem using the standard distributions which i'm missing? or have i done this correctly (at least to the authors intentions) and figured out a way of generating it myself?

further is there any "tricks" or guiding factors which make you strongly consider one distribution over another other than the "random with big sample sizes, fairly low dependence.....poisson" type of thinking?

any and all clarification or help would be greatly appreciated. i'm just a little worried that i've not actually fully grasped the chapter and this will hinder me later on in the book.

• Not every probability distribution you encounter in life will be one of the special ones you find in a textbook. But if you must, you can think of this as the uniform distribution over all permutations of $10$ people, i.e. each outcome has equal probability $1/10!$. Then computing probabilities of an event amounts to counting how many outcomes lie in that event and multiplying by $1/10!$, which you have done. – angryavian Jul 28 '17 at 0:04
• i understand that not every probability distribution i come across will fit neatly into the small collection. i guess i just assumed (perhap wrongly) that the chapter problems following the chapter summary would be inclined towards re-affirming the core principles of the chapter. which is why i thought i'd ask. still i found your comment helpful, i hadnt considered a uniform distribution. so thank you. – Vaas Jul 28 '17 at 0:10
• The core principle is: how to count arrangements and derive a probability mass function from first principles. – Graham Kemp Jul 28 '17 at 0:50
• @Graham Kemp ahhh i see. So in all sincerity it was me misunderstanding what the core principle of the chapter was and put too much emphasis on the basic distributions themselves....also thank you for help clearing that up. Ill go reread the chapter with this in mind. – Vaas Jul 28 '17 at 0:54

You found the probability for selecting $i-1$ from $5$ men and $1$ from $5$ women, when selecting groups of size $i-1$ and $1$ from any of the $10$ men-and-women, without bias or replacement.   As required.
\begin{align*}\mathsf P\{X=i\}~&=~\dfrac{\dbinom{5}{i-1}\dbinom{5}1}{\dbinom{10}{i-1,1,10-i}} \\[2ex] &=~\dfrac{5!~5~(10-i)!}{(6-i)!~10!}\end{align*}