Question

Two fair dice are tossed, and the face on each die is observed.

1. Use a tree diagram to find the 36 sample points contained in the sample space.
2. Assign probabilities to the sample points in part a.
3. Find the probability of each of the following events:

$\text{A = {}3$showing on each die}

$\text{B = {}$Sum of two numbers showing is}

$\text{C = {}$Sum of two numbers showing is even}

Short answer

1. Tree Diagram
2. Table
3. $\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{36}}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{C}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Step-by-step solution

Finding the sample points contained in sample space

Two dice are tossed and the face on each dice is observed. The first dice is tossed and the face on the dice is written. After that, the second dice is tossed and the face on this dice is observed. The first branches show the face on the first dice and the ends show the face on the second dice.

Assigning probability to sample points

Two fair dice are tossed, the faces on one dice are 1, 2, 3, 4, 5 and 6 .

Therefore, the total number of possible outcomes are ${\mathbf{6}}^{\mathbf{2}}\mathbf{=}\mathbf{36}$

Finding the probability of event A,B and C

$\mathbf{A}\mathbf{=}\mathbf{3}$showing on each die and$\mathbf{S}\mathbf{=}$total outcome (sample space)

One observes 3 on both the dice only one time.

Hence $\mathbf{n}\left(\mathbf{B}\right)\mathbf{=}\mathbf{1}$ and $\mathbf{n}\left(\mathbf{S}\right)\mathbf{=}\mathbf{36}$

localid="1662212210908" $\begin{array}{rcl}\mathbf{P}\left(\mathbf{A}\right)& \mathbf{=}& \frac{\mathbf{Favorable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}\\ & \mathbf{=}& \frac{\mathbf{n}\left(\mathbf{A}\right)}{\mathbf{n}\left(\mathbf{S}\right)}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{36}}\end{array}$

Therefore, the probability of getting 3 on each die is localid="1662212168718" $\frac{\mathbf{1}}{\mathbf{36}}$.

$\mathbf{B}\mathbf{=}$Sum of two numbers showing is 7

One observes sum of two numbers showing as 7 when the sample points are $\left(\mathbf{1}\mathbf{,}\mathbf{6}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{5}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{5}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{6}\mathbf{,}\mathbf{1}\right)$

Therefore, $\mathbf{n}\left(\mathbf{B}\right)\mathbf{=}\mathbf{6}$

localid="1662212248503" $\begin{array}{rcl}\mathbf{P}\left(\mathbf{B}\right)& \mathbf{=}& \frac{\mathbf{Favorable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{out}\mathbf{}\mathbf{comes}}\\ & \mathbf{=}& \frac{\mathbf{n}\left(\mathbf{B}\right)}{\mathbf{n}\left(\mathbf{S}\right)}\\ & \mathbf{=}& \frac{\mathbf{6}}{\mathbf{36}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{6}}\end{array}$

Thus, the probability of getting two numbers whose sum is 7 is$\frac{\mathbf{1}}{\mathbf{6}}$.

$\mathbf{C}\mathbf{=}$Sum of two numbers is even

One observes sum of two numbers is even when the sample points are

$$\begin{gathered} \left(\mathbf{1}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{1}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{1}\mathbf{,}\mathbf{5}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{6}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{5}\right) \\ \left(\mathbf{4}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{6}\right)\mathbf{,}\left(\mathbf{5}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{5}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{5}\mathbf{,}\mathbf{5}\right)\mathbf{,}\left(\mathbf{6}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{6}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{6}\mathbf{,}\mathbf{6}\right) \end{gathered}$$

Therefore, $\mathbf{n}\left(\mathbf{C}\right)\mathbf{=}\mathbf{18}$

localid="1662212281500" $\begin{array}{rcl}\mathbf{P}\left(\mathbf{C}\right)& \mathbf{=}& \frac{\mathbf{Favorable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}\\ & \mathbf{=}& \frac{\mathbf{n}\left(\mathbf{C}\right)}{\mathbf{n}\left(\mathbf{S}\right)}\\ & \mathbf{=}& \frac{\mathbf{18}}{\mathbf{36}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\end{array}$

Hence, the probability of getting two numbers whose sum is an even number is $\frac{\mathbf{1}}{\mathbf{2}}$.