Chapter 5.3, Problem 7P

Tree Diagram There are six balls in an urn. They are identical except for color. Two are red, three are blue, and one is yellow. You are to draw a ball from the urn, note its color, and set it aside. Then you are to draw another ball from the urn and note its color.

(a) Make a tree diagram to show all possible outcomes of the experiment. Label the probability associated with each stage of the experiment on the appropriate branch.

(b) Probability Extension Compute the probability for each outcome of the experiment.

To determine

To draw: A tree diagram.

Explanation of Solution

Given: The number of red, blue and yellow balls are 2,3, and 1, respectively.

Graph: A tree diagram of the provided experiment is shown below:

Interpretation: In the tree diagram, initially three branches have been drawn which indicate the three possible choices of selecting red, blue and yellow balls. Thereafter, there are three branches for each ball which represent the possible outcomes of selecting the individual balls. At the end of the tree diagram, there are a total of 9 branches which show the total number of possible outcomes.

To determine

The probabilities for each outcome of the provided experiment.

Answer to Problem 7P

Solution: The required probabilities are $P\left(R,R\right)=\frac{1}{15}$, $P\left(B,R\right)=\frac{1}{5}$, $P\left(Y,R\right)=\frac{1}{15}$, $P\left(R,B\right)=\frac{1}{5}$, $P\left(B,B\right)=\frac{1}{5}$, $P\left(Y,B\right)=\frac{1}{10}$, $P\left(R,Y\right)=\frac{1}{15}$, and $P\left(B,Y\right)=\frac{1}{10}$.

Explanation of Solution

Given: The total number of balls are 6, out of which, two are red, three are blue and one is yellow.

Consider A and B to be the two events. The formula to calculate the probability of A and B occurring together using the multiplication rule is:

$P\left(A\cap B\right)=P\left(A|B\right)\times P\left(B\right)$

From the tree diagram, following are the possible outcomes of the experiment:

$\begin{array}{l}RR=\text{}\text{}\text{ Red on 1st and red on 2nd}\\ RB=\text{Red on 1st and blue on 2nd}\\ RY=\text{Red on 1st and yellow on 2nd}\end{array}$

Further,

$\begin{array}{l}BR=\text{Blue on 1st and red on 2nd}\\ BB=\text{Blue on 1st and blue on 2nd}\\ BY=\text{Blue on 1st and yellow on 2nd}\end{array}$

The possible outcomes of the event are,

$\begin{array}{l}YR=\text{Yellow on 1st and red on 2nd}\\ YB=\text{Yellow on 1st and blue on 2nd}\end{array}$

Here, the multiplication rule will be used for dependent events to compute the probability of each outcome. The required probabilities are:

The probability of $P\left(R,R\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(R,R\right)=P\left(R\right)\times P\left(R|R\right)\\ =\frac{2}{6}\times \frac{1}{5}\\ =\frac{1}{15}\end{array}$

The probability of $P\left(R,B\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(R,B\right)=P\left(R\right)\times P\left(B|R\right)\\ =\frac{2}{6}\times \frac{3}{5}\\ =\frac{1}{5}\end{array}$

The probability of $P\left(R,Y\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(R,Y\right)=P\left(R\right)\times P\left(Y|R\right)\\ =\frac{2}{6}\times \frac{1}{5}\\ =\frac{1}{15}\end{array}$

The probability of $P\left(B,R\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(B,R\right)=P\left(B\right)\times P\left(R|B\right)\\ =\frac{3}{6}\times \frac{2}{5}\\ =\frac{1}{5}\end{array}$

The probability of $P\left(B,B\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(B,B\right)=P\left(B\right)\times P\left(B|B\right)\\ =\frac{3}{6}\times \frac{2}{5}\\ =\frac{1}{5}\end{array}$

The probability of $P\left(B,Y\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(B,Y\right)=P\left(B\right)\times P\left(Y|B\right)\\ =\frac{3}{6}\times \frac{1}{5}\\ =\frac{1}{10}\end{array}$

The probability of $P\left(Y,R\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(Y,R\right)=P\left(Y\right)\times P\left(R|Y\right)\\ =\frac{1}{6}\times \frac{2}{5}\\ =\frac{1}{15}\end{array}$

The probability of $P\left(Y,B\right)$ can be calculated as:

$\begin{array}{c}P\text{}\text{}\left(Y,B\right)=P\left(Y\right)\times P\left(B|Y\right)\\ =\frac{1}{6}\times \frac{3}{5}\\ =\frac{1}{10}\end{array}$

Hence, the required probabilities are $P\left(R,R\right)=\frac{1}{15}$, $P\left(B,R\right)=\frac{1}{5}$, $P\left(Y,R\right)=\frac{1}{15}$, $P\left(R,B\right)=\frac{1}{5}$, $P\left(B,B\right)=\frac{1}{5}$, $P\left(Y,B\right)=\frac{1}{10}$, $P\left(R,Y\right)=\frac{1}{15}$, and $P\left(B,Y\right)=\frac{1}{10}$.