Probability

Of $27$ students in a class, $11$ have blue eyes, $13$ have brown eyes, and $3$ have green eyes. If $3$ students are chosen at random, what are the odds of each event occurring?

1. $2$ have brown eyes and $1$ has blue eyes
2. only $1$ has green eyes

$3$ students can be selected from $27$ students in

${^{27}}{P}_{3} = \dfrac{27!}{24!} = 27 \times 26 \times 25 = 17,550$ ways

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 17550$

1. Let $A$ be the event of selecting $2$ students who have brown eyes AND $1$ student who has blue eyes
   
   $2$ students with brown eyes can be selected from $13$ students in
   
   ${^{13}}{P}_{2} = \dfrac{13!}{11!} = 13 \times 12 = 156$ ways
   
   $1$ student with blue eyes can be selected from $11$ students in $11$ ways
   
   Amongst themselves, the $3$ students can be selected in $3$ ways
   
   $\therefore \;$ Number of ways of selecting $2$ students with brown eyes and $1$ student with blue eyes
   
   $= 156 \times 11 \times 3 = 5148$ ways
   
   $\therefore \;$ Number of elements in $A = n\left(A\right) = 5148$
   
   Let $P \left(s\right)$ be the probability of success of event $A$ and $P \left(f\right)$ be the probability of failure of event $A$
   
   $P \left(s\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{5148}{17550} = \dfrac{286}{975}$
   
   $P \left(f\right) = 1- \dfrac{286}{975} = \dfrac{689}{975}$
   
   $\therefore \;$ Odds of event $A = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{286/975}{689/975} = \dfrac{22}{53}$
   
   
2. Let $B$ be the event of selecting $3$ students such that only $1$ student has green eyes
   
   $1$ student with green eyes can be selected from $3$ students in ${^{3}}{P}_{1} = 3$ ways
   
   Remaining $2$ students can be selected from the remaining $24$ students in
   
   ${^{24}}{P}_{2} = \dfrac{24!}{22!} = 24 \times 23$ ways
   
   Amongst themselves, these $3$ students can be selected in $3$ ways
   
   $\therefore \;$ Number of ways of selecting $3$ students such that only $1$ student has green eyes
   
   $= 3 \times 24 \times \times 23 \times 3 = 4968$ ways
   
   $\therefore \;$ Number of elements in $B = n\left(B\right) = 4968$
   
   Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$
   
   $P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{4968}{17550} = \dfrac{92}{325}$
   
   $P \left(f\right) = 1- \dfrac{92}{325} = \dfrac{233}{325}$
   
   $\therefore \;$ Odds of event $B = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{92/325}{233/325} = \dfrac{92}{233}$