Chapter 11.9, Problem 11E

a.

To determine

To find the probability of a gumball machine dispensing two red gumballs in a row.

Answer to Problem 11E

The probability of a gumball machine dispensing two red gumballs in a row is $\frac{7}{59}$ .

Explanation of Solution

Given information:

The gumball machine has $21$ red gumballs, $24$ yellow gumballs and $15$ blue gumballs.

Calculation :

The gumball machine dispensing two gumballs in a row is a probability of dependent events.

Probability of dispensing a red gumball, $\text{P}\left(\text{red}\right)\text{=}\frac{21}{60}$

Probability of dispensing another red gumball, $\text{P}\left(\text{another red}\right)\text{=}\frac{20}{59}$

  $\begin{array}{l}\text{P}\left(\text{red and red}\right)\text{=P}\left(\text{red}\right)\cdot \text{P}\left(\text{another red}\right)\\ \text{P}\left(\text{red and red}\right)=\frac{21}{60}\cdot \frac{20}{59}\left[\text{Substitute probabilities}\right]\\ \text{P}\left(\text{red and red}\right)=\frac{7}{59}\left[\text{Multiply}\right]\end{array}$

Hence, the probability of a gumball machine dispensing two red gumballs in a row is $\frac{7}{59}$ .

b.

To determine

To compare the probability that the gumball machine dispenses red gumball, then a yellow gumball and the probability that the machine dispenses a red gumball, then a blue gumball.

Answer to Problem 11E

The probability that the machine dispenses a red, then a yellow gumball is greater.

Explanation of Solution

Given information:

The gumball machine has $21$ red gumballs, $24$ yellow gumballs and $15$ blue gumballs.

Calculation :

The gumball machine dispensing two gumballs in a row is a probability of dependent events.

Probability of dispensing a red gumball, $\text{P}\left(\text{red}\right)\text{=}\frac{21}{60}$

Probability of dispensing another yellow gumball, $\text{P}\left(\text{yellow}\right)\text{=}\frac{24}{59}$

  $\begin{array}{l}\text{P}\left(\text{red and yellow}\right)\text{=P}\left(\text{red}\right)\cdot \text{P}\left(\text{yellow}\right)\\ \text{P}\left(\text{red and yellow}\right)=\frac{21}{60}\cdot \frac{24}{59}\left[\text{Substitute probabilities}\right]\\ \text{P}\left(\text{red and yellow}\right)=\frac{42}{295}\left[\text{Multiply}\right]\end{array}$

Hence, the probability of a gumball machine dispensing a red gumball, then a yellow gumballis $\frac{42}{295}\text{ or 0}\text{.142}$ .

Probability of dispensing a red gumball, $\text{P}\left(\text{red}\right)\text{=}\frac{21}{60}$

Probability of dispensing another blue gumball, $\text{P}\left(\text{blue}\right)\text{=}\frac{15}{59}$

  $\begin{array}{l}\text{P}\left(\text{red and blue}\right)\text{=P}\left(\text{red}\right)\cdot \text{P}\left(\text{blue}\right)\\ \text{P}\left(\text{red and blue}\right)=\frac{21}{60}\cdot \frac{15}{59}\left[\text{Substitute probabilities}\right]\\ \text{P}\left(\text{red and blue}\right)=\frac{21}{236}\left[\text{Multiply}\right]\end{array}$

Hence, the probability of a gumball machine dispensing a red gumball, then a blue gumball is $\frac{21}{236}\text{ or 0}\text{.089}$ .

The probability that the machine dispenses a red, then a yellow gumball is greater.