Chapter 12.4, Problem 3CYP

To determine

Tofind:the probability that she pullsout four of one color and three of another after adding seven marble to the bag.

Answer to Problem 3CYP

The probability of pulling $4$ of one color and $3$ of another colors is $0.07547$ .

Explanation of Solution

Given:

  $7$ green marble is added to the bag.

Concept used:

The probability can be finds out through using permutation and combination as:

If order is matter than permutation is used.

  $\mathrm{P}{}_{\mathrm{r}}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}$ and $\mathrm{C}{}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}$ .

  $\mathrm{C}{}_{\mathrm{r}}=\frac{1}{\mathrm{r}!}\mathrm{P}{}_{\mathrm{r}}\text{ or }\mathrm{r}!\mathrm{C}{}_{\mathrm{r}}=\mathrm{P}{}_{\mathrm{r}}.$

Calculation:

According to the given $7$ green marble is added to the bag.

Now total marble is $35$ .

If order is matter $2$ colors are chosen from the five different colors as:

  $\mathrm{P}{}_2=\frac{5!}{3!}=20.$

The probability of getting $4$ marble of same colors chosen from a group of $7$ is:

  $\mathrm{C}{}_4=\frac{7!}{4!\left(3!\right)}=35.$

The probability of getting a $3$ another color chosen from a group of $7$ .

  $\mathrm{C}{}_3=\frac{7!}{3!4!}=35.$

Use permutations and combinations along with the fundamental counting principle, to find the probability of success which is denoted by $\mathrm{s}$ .

  $\begin{array}{l}\mathrm{P}{}_2\times \mathrm{C}{}_4\times \mathrm{C}{}_3=20\times 35\times 35.\\ \Rightarrow 24500.\end{array}$

Determining the number of possibilities is probability of success plus failure $\left(\mathrm{s}+\mathrm{f}\right)$ :

  $\mathrm{C}{}_5=324632.$

The probability of getting $4$ of one color and $3$ of another colors is:

  $\begin{array}{l}\frac{\mathrm{s}}{\mathrm{s}+\mathrm{f}}=\frac{24500}{324632}.\\ \Rightarrow \mathrm{0.07547.}\end{array}$

Hence, the probability of pulling $4$ of one color and $3$ of another colors is $0.07547$ .