Chapter 13, Problem 34SGA

To determine

To find: The odds of each occurring event.

Answer to Problem 34SGA

The odds to get one is nickel and two are dimes is $\frac{1}{13}$ .

Explanation of Solution

Given:

The bag containing $7$ pennies, $4$ nickels, and $5$ dimes.

The given event: Three coins are drawn at random, of them one is nickel and two are dimes.

Calculation:

The probability of getting one is nickel and two are dimes is calculated as:

  $\begin{array}{c}P\left(E\right)=\frac{C\left(4,1\right)\cdot C\left(5,2\right)}{C\left(16,3\right)}\left[\begin{array}{l}C\left(4,1\right)\text{is the number of ways of selecting one nickel from the bag}\\ C\left(5,2\right)\text{is the number of ways of selecting two dimes from the bag}\\ C\left(16,3\right)\text{is the number of ways of selecting three coins from the bag}\end{array}\right]\\ =\frac{\frac{4!}{\left(4-1\right)!1!}\cdot \frac{5!}{\left(5-2\right)!2!}}{\frac{16!}{\left(16-3\right)!3!}}\\ =\frac{\frac{4!}{1!3!}\cdot \frac{5!}{\left(3\right)!2!}}{\frac{16!}{13!3!}}\\ =\frac{4!\times 13!\times 5!}{16!\times 2!\times 3!}\end{array}$

Further simplified as:

  $P\left(E\right)=\frac{1}{14}$ .

The probability of not getting three pennies is calculated as:

  $\begin{array}{c}P\left(E^{\prime }\right)=1-P\left(E\right)\left[P\left(E^{\prime }\right)\text{is probability of failure}\right]\\ =1-\frac{1}{14}\\ =\frac{13}{14}\end{array}$

We know that the odds of the successful outcome of an event is the ratio of the probability of its success to the probability of its failure.

  $\begin{array}{c}\text{Odds}=\frac{P\left(E\right)}{P\left(E^{\prime }\right)}\\ =\frac{\frac{1}{14}}{\frac{13}{14}}\\ =\frac{1}{13}\end{array}$

Therefore the odds to get one is nickel and two are dimesis $\frac{1}{13}$ .