91Ó°ÊÓ

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.

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 ${\mathbf{6}}^{\mathbf{2}}\mathbf{=}\mathbf{36}$

$\mathbf{A}\mathbf{=}\mathbf{3}$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 $\frac{\mathbf{1}}{\mathbf{2}}$.