Chapter 13.3, Problem 28E

To determine

To find: The odds of drawing 1 green and 1 red marbles.

Answer to Problem 28E

The odds of drawing 1 green and 1 red marbles is $\frac{11}{4}$ .

Explanation of Solution

The number of green marbles are $1$ .

The number of yellow marbles are $2$ .

The number of red marbles are $3$ .

The probability of the successful outcome of an event is given by,

  $P\left(s\right)=\frac{s}{s+f}$

The probability of drawing 1 green and 1 redmarble is calculated as,

  $\begin{array}{c}P\left(\text{s}\right)=\frac{11}{C\left(6,2\right)}\\ =\frac{11}{15}\end{array}$

The probability of failure is calculated as,

  $\begin{array}{c}P\left(f\right)=1-P\left(s\right)\\ =1-\frac{11}{15}\\ =\frac{4}{15}\end{array}$

The odds of drawing 1 green and 1 redis calculated as,

  $\begin{array}{c}\text{Odds}=\frac{P\left(s\right)}{P\left(f\right)}\\ =\frac{11/15}{4/15}\\ =\frac{11}{4}\end{array}$

Therefore, the odds of drawing 1 green and 1 redmarbleis $\frac{11}{4}$ .