Probability

A box contains $1$ green, $2$ yellow and $3$ red marbles. Two marbles are drawn at random without replacement. What are the odds of each event occurring?

1. not drawing yellow marbles;
2. drawing marbles of two different colors.

$2$ marbles can be drawn from $1 + 2 + 3 = 6$ marbles (without replacement) in

${^{6}}{P}_{2} = \dfrac{6!}{4!} = 6 \times 5 = 30$ ways

$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 30$

1. Let $A$ be the event of not drawing yellow marbles
   
   Then $2$ marbles can be drawn from the remaining $4$ marbles in ${^{4}}{P}_{2} = \dfrac{4!}{2!} = 12$ ways
   
   $\therefore \;$ Number of elements in $A = n \left(A\right) = 12$
   
   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 $A$
   
   $P \left(s\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{12}{30} = \dfrac{2}{5}$
   
   $P \left(f\right) = 1- \dfrac{2}{5} = \dfrac{3}{5}$
   
   $\therefore \;$ Odds of event $A = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{2 / 5}{3 / 5} = \dfrac{2}{3}$
   
   
2. Let $B$ be the event of drawing marbles of two different colors
   
   Then, the possibilities for event $B$ are
   
   ($1$ green AND $1$ yellow) OR ($1$ green AND $1$ red) OR ($1$ yellow AND $1$ red)
   
   $1$ green AND $1$ yellow marble:
   
   $1$ green marble can be selected in $1$ way
   
   $1$ yellow marble can be selected from $2$ yellow marbles in $2$ ways
   
   Amongst themselves, $1$ green AND $1$ yellow marble can be selected in in $2$ ways
   
   $\therefore \;$ $1$ green AND $1$ yellow marble can be selected in $1 \times 2 \times 2 = 4$ ways
   
   $1$ green AND $1$ red marble:
   
   $1$ green marble can be selected in $1$ way
   
   $1$ red marble can be selected from $3$ red marbles in $3$ ways
   
   Amongst themselves, $1$ green AND $1$ red marble can be selected in in $2$ ways
   
   $\therefore \;$ $1$ green AND $1$ red marble can be selected in $1 \times 3 \times 2 = 6$ ways
   
   $1$ yellow AND $1$ red marble:
   
   $1$ yellow marble can be selected from $2$ yellow marbles in $2$ ways
   
   $1$ red marble can be selected from $3$ red marbles in $3$ ways
   
   Amongst themselves, $1$ yellow AND $1$ red marble can be selected in in $2$ ways
   
   $\therefore \;$ $1$ green AND $1$ red marble can be selected in $2 \times 3 \times 2 = 12$ ways
   
   $\therefore \;$ Two marbles of two different colors can be selected in $4 + 6 + 12 = 22$ ways
   
   $\therefore \;$ Number of elements in $B = n \left(B\right) = 22$
   
   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 $B$
   
   $P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{22}{30} = \dfrac{11}{15}$
   
   $P \left(f\right) = 1- \dfrac{11}{15} = \dfrac{4}{15}$
   
   $\therefore \;$ Odds of event $B = \dfrac{P \left(s\right)}{P \left(f\right)} = \dfrac{11 / 15}{4 / 15} = \dfrac{11}{4}$