Question

From a bag containing $5$ nickels, $8$ dimes, and $7$ quarters, $5$ coins are drawn at random and all at once. What is the probability of getting $2$ nickels, $2$ dimes, and $1$ quarter?

Answer 1

Step-1: Find the total number of possible outcomes:

Given that the bag containing $5$ nickels, $8$ dimes, and $7$ quarters, $5$ coins are drawn at random and all at once

Use the probability formula:

$\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$

Use the combination formula ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\times (\mathrm{n}-\mathrm{r})!}$ where $\mathrm{n}$ represents the number of items, and $\mathrm{r}$ represents the number of items being chosen at a time.

Here, the total number of items are:

$\begin{array}{rcl}& \Rightarrow & \mathrm{n}=5+8+7\\ & =& 20\end{array}$

Five coins are chosen at random and drawn from the bag all at once:

$\begin{array}{rcl}& \Rightarrow & \mathrm{r}=5\end{array}$

Substitute the known values in the formula ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\times (\mathrm{n}-\mathrm{r})!}$:

$\begin{array}{rcl}& \Rightarrow & {}^{20}{\mathrm{C}}_5=\frac{20!}{5!\times (20-5)!}\\ & =& \frac{20\times 19\times 18\times 17\times 16\times 15!}{5!\times 15!}\\ & =& \frac{20\times 19\times 18\times 17\times 16}{5!}[\mathrm{Cancel}\mathrm{the}\mathrm{common}\mathrm{term}]\\ & =& \frac{20\times 19\times 18\times 17\times 16}{5\times 4\times 3\times 2\times 1}\\ & =& 15504\end{array}$

Step-2: Find the number of favorable outcomes:

The number of nickels is $5$. Choosing $2$ nickels from these $5$ nickels, then the number of cases is ${}^5{\mathrm{C}}_2$ ways.

The number of dimes is $8$. Choosing $2$ dimes from these $8$ dimes, then the number of cases is ${}^8{\mathrm{C}}_2$ ways.

The number of quarters is $7$. Choosing $1$ quarter from these $7$ quarters, then the number of cases is ${}^7{\mathrm{C}}_1$ways.

Therefore, the favorable number of cases is:

$\begin{array}{rl}\mathrm{Favourable}\mathrm{outcomes}& {=}^5{\mathrm{C}}_2{\times }^8{\mathrm{C}}_2{\times }^7{\mathrm{C}}_1\\ & =\frac{5!}{2!(5-2!)}\times \frac{8!}{2!(8-2)!}\times \frac{7!}{1!(7-1)!}\\ & =\frac{5!}{2!\times 3!}\times \frac{8!}{2!\times 6}\times \frac{7!}{1!\times 6!}\\ & =\frac{5\times 4\times 3\times 2\times 1!}{2\times 3\times 2\times 1!}\times \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1!}{2\times 6\times 5\times 4\times 3\times 2\times 1!}\times \frac{7\times 6\times 5\times 4\times 3\times 2\times 1!}{1\times 6\times 5\times 4\times 3\times 2\times 1!}\\ & =\frac{120}{12}\times \frac{40320}{2\times 720}\times \frac{5040}{720}\\ & =\frac{3360}{12}\times \frac{5040}{720}\\ & =\frac{5880}{3}\\ & =1960\end{array}$

Step-3: Find the probability of getting $\mathbf{2}$ nickels, $\mathbf{2}$ dimes, and $\mathbf{1}$ quarter:

Substitute the total number of outcomes $=15504$ and possible outcomes $=1960$ into the $\mathrm{P}\left(\mathrm{A}\right)$ formula:$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}\\ & =& \frac{1960}{15504}\\ & =& 0.1264\end{array}$

Hence, the probability of getting $\mathbf{2}$ nickels, $\mathbf{2}$ dimes, and $\mathbf{1}$ quarter is $\mathbf{0}\mathbf{.}\mathbf{1264}.$