Q.4.4 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

$P\left(X\right)=\{1/2,5/18,5/36,5/84,5/252,1/252,0,0,0,0\}$ for $x=1,2,3...,10$ respectively.

Step 1:Given Information

The total number of paths of ranking 10 different scores is 10!. The number of ways a female can be ranked 1 is.

ways of choosing any one out of 5. ways of arranging the rest 9

$P(X=1)=\frac{{}^5{C}_1{\cdot }^9{P}_9}{{}^{10}{P}_{10}}=\frac{5.9!}{10!}=\frac{1}{2}$

Step 2:Explanation

The number of ways a female can be ranked 2 is,

Ways of female can be ranked 2 or less: ways that Ist male and female/5.Ways rest 8 are ranked.

$P(X=2)=\frac{{}^5{P}_1{\cdot }^5{C}_1{\cdot }^8{P}_8}{{}^{10}{P}_{10}}=\frac{\mathrm{5.5.8}!}{10!}=\frac{5}{18}$

Step 3:Explanation

$P(X=3)=\frac{{}^5{P}_2{\cdot }^5{C}_1{\cdot }^7{P}_7}{{}^{10}{P}_{10}}=\frac{\mathrm{20.5.7}!}{10!}=\frac{5}{36}$

Step 4:Explanation

$P(X=4)=\frac{{}^5{P}_3{\cdot }^5{C}_1{\cdot }^6{P}_6}{{}^{10}{P}_{10}}=\frac{60\cdot 5\cdot 6!}{10!}=\frac{5}{84}$

Step 5:Explanation

$P(X=5)=\frac{{}^5{P}_4{\cdot }^5{C}_1{\cdot }^5{P}_5}{{}^{10}{P}_{10}}=\frac{\mathrm{120.5.5}!}{10!}=\frac{5}{252}$

Step 6:Explanation

$P(X=6)=\frac{{}^5{P}_5{\cdot }^5{C}_1{\cdot }^4{P}_4}{{}^{10}{P}_{10}}=\frac{120\cdot 5\cdot 4!}{10!}=\frac{1}{252}$

Step 7:Explanation

There existing only$5$boys, the lower value of $X$ can be $6$

$P(X>6)=0$

Step 8:Final Answer

$P\left(X\right)=\{1/2,5/18,5/36,5/84,5/252,1/252,0,0,0,0\}$ for $x=1,2,3,...10$ respectively.