91影视

The diagram depicted in the question is a Venn diagram. The diagram uses circles to show relationships among various things. Here, one circle represents sample points in event A and the other represents sample points in event B.

Given that the sample points are all equally likely,

$\mathbf{P}\left(\mathbf{1}\right)\mathbf{=}\mathbf{P}\left(\mathbf{2}\right)\mathbf{=}\mathbf{P}\left(\mathbf{3}\right)\mathbf{=}\mathbf{P}\left(\mathbf{4}\right)\mathbf{=}\mathbf{P}\left(\mathbf{5}\right)\mathbf{=}\mathbf{P}\left(\mathbf{6}\right)\mathbf{=}\mathbf{P}\left(\mathbf{7}\right)\mathbf{=}\mathbf{P}\left(\mathbf{8}\right)\mathbf{=}\mathbf{P}\left(\mathbf{9}\right)\mathbf{=}\mathbf{P}\left(\mathbf{10}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}$

Sample points lying in event A are 4, 5 and 6.

$$\begin{gathered} \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{P}\left(\mathbf{4}\right)\mathbf{+}\mathbf{P}\left(\mathbf{5}\right)\mathbf{+}\mathbf{P}\left(\mathbf{6}\right) \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}} \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}} \end{gathered}$$

Sample points lying in event B are 5, 6, 7 and 8.

$$\begin{gathered} \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{P}\left(\mathbf{5}\right)\mathbf{+}\mathbf{P}\left(\mathbf{6}\right)\mathbf{+}\mathbf{P}\left(\mathbf{7}\right)\mathbf{+}\mathbf{P}\left(\mathbf{8}\right) \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}} \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{10}} \end{gathered}$$

Therefore, the $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}$ and $\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{10}}$.

Given that,

$\mathbf{P}\left(\mathbf{1}\right)\mathbf{=}\mathbf{P}\left(\mathbf{2}\right)\mathbf{=}\mathbf{P}\left(\mathbf{3}\right)\mathbf{=}\mathbf{P}\left(\mathbf{4}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{16}}$ and $\mathbf{P}\left(\mathbf{5}\right)\mathbf{=}\mathbf{P}\left(\mathbf{6}\right)\mathbf{=}\mathbf{P}\left(\mathbf{7}\right)\mathbf{=}\mathbf{P}\left(\mathbf{8}\right)\mathbf{=}\mathbf{P}\left(\mathbf{9}\right)\mathbf{=}\mathbf{P}\left(\mathbf{10}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}$

Sample points lying in event A are 4, 5 and 6:

$$\begin{gathered} \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{P}\left(\mathbf{4}\right)\mathbf{+}\mathbf{P}\left(\mathbf{5}\right)\mathbf{+}\mathbf{P}\left(\mathbf{6}\right) \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}} \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{5}}{\mathbf{16}} \end{gathered}$$

Sample points lying in event B are 5, 6, 7 and 8:

$$\begin{gathered} \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{P}\left(\mathbf{5}\right)\mathbf{+}\mathbf{P}\left(\mathbf{6}\right)\mathbf{+}\mathbf{P}\left(\mathbf{7}\right)\mathbf{+}\mathbf{P}\left(\mathbf{8}\right) \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}} \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{8}} \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Therefore, the$\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{5}}{\mathbf{16}}$and$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$.