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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 $P\{X=i\}$, $i=1,2,3,...,8,9,10$.

I got p(1) and p(2) right. For p(3) I have $p(3) = \frac{\binom{5}{2}5*7!}{10!} = \frac{5}{72}$ but apparently, the correct answer is $\frac{5*4*5}{10*9*8}$.

The reasoning behind my answer is that because there are 10! possible combinations, we find the possible combinations for the first 3 being men, followed by a women, and the rest in whatever order, and divide this quantity by 10!. Thus having on the top portion being (5 choose 2)*5*6! What did I do wrong?

Any help would be appreciated, L

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    $\begingroup$ Your $\binom{5}{2}$ counts the number of ways to choose the $2$ men who will be top ranked, but not the order in which they will come. So instead of $\binom{5}{2}$ you need $2!\binom{5}{2}$, that is, $(5)(4)$. $\endgroup$ – André Nicolas May 5 '15 at 0:57
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Here we use the multiplication-principle (probability version).

The multiplication principle

$$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$$

You can show by induction then that for example $Pr(A\cap B\cap C) = Pr(A)\cdot Pr(B|A)\cdot Pr(C|A\cap B)$

Let $A_m, B_m, C_w$ denote that the top ranked individual is a man, the second ranked individual is a man, and the third ranked individual is a woman respectively.

Then $Pr(X=3) = Pr(A_m\cap B_m\cap C_w)$.

Complete the problem by applying the multiplication principle.


For calculating $Pr(X=1)=Pr(A_w)$ that would be $\frac{5}{10}$ since there are 5 women out of 10 people total.

For calculating $Pr(X=2)=Pr(A_m\cap B_w)=Pr(A_m)\cdot Pr(B_w|A_m) = \frac{5}{10}\cdot \frac{5}{9}$ since originally there are 5 men out of 10 people total, followed by 5 women out of 9 people total remaining.

As such, $Pr(X=3) = Pr(A_m\cap B_m\cap C_w) = Pr(A_m)\cdot Pr(B_m|A_m)\cdot Pr(C_w|A_m\cap B_m)$

$ = \frac{5}{10}\cdot \frac{4}{9}\cdot \frac{5}{8}=\frac{5\cdot 4\cdot 5}{10\cdot 9\cdot 8}$


As noted by Andre above (just as I was writing my own version of the same comment), your proposed solution mistakenly uses $\binom{5}{2}$ where it should have been $5\cdot 4$. Approaching this from a counting perspective instead of a probability perspective, we have:

$$\underbrace{\underline{~~}\underline{~~}}_{2men}~\underbrace{\underline{~~}}_{woman}\underline{~~}\underline{~~}\cdots$$

Pick which man takes the first spot: 5 choices

Pick which man takes the second spot: 4 choices

Pick which woman takes the third spot: 5 choices

Pick how to arrange the remaining 7 people: 7! choices

$$\underbrace{\underline{~~}}_{5}\underbrace{\underline{~~}}_{4}~\underbrace{\underline{~~}}_{5}\underbrace{\underline{~~}\underline{~~}\cdots}_{7!}$$

As such there are $5\cdot 4\cdot 5\cdot 7!$ different such arrangements out of $10!$ arrangements total, for a probability of $Pr(X=3)=\frac{5\cdot 4\cdot 5\cdot 7!}{10!} = \frac{5\cdot 4\cdot 5}{10\cdot 9\cdot 8}$

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  • $\begingroup$ Is it a bad idea to approach it using combinatorics? ;o $\endgroup$ – leixir May 5 '15 at 1:01
  • $\begingroup$ @leixir the multiplication principle (probability version) works almost exactly the same as the multiplication principle (counting version). The answers are equivalent, though using the probability version lets you avoid needing to pay attention to the remaining $7!$ on the top and bottom of the fraction (since we only focus on the first so many terms). $\endgroup$ – JMoravitz May 5 '15 at 1:04
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    $\begingroup$ Not a bad idea, but here I prefer direct probability calculation. The probability that the first is a man is $\frac{5}{10}$. Given that the first is a man, the probability the second is a man is $\frac{4}{9}$. Given the first two were men, the probability the third is a woman is $\frac{5}{8}$. Multiply. $\endgroup$ – André Nicolas May 5 '15 at 1:07
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You might be confusing combination and permutation in your analysis. the question does not consider all 10! permutations, it is only considering 10*9*8 of them. There are 10 possibilities for the first choice, 9 for the second, and 8 for the third. So with the question, what is the probability that a woman has the third highest score, there are 720 possible outcomes for any order if there are 10 people and 3 ranks. That is the total sample space for that problem. For the numerator, which represents the number of permutations where a woman is third, there are 5*4*5 by the same logic.

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