Question

The diagram below describes the sample space of a particular experiment and events A and B .

1. What is this type of diagram called?
2. Suppose the sample points are equally likely. Find $\mathbf{P}\left(\mathbf{A}\right)$and $\mathbf{P}\left(\mathbf{B}\right)$.
3. Suppose $\mathbf{P}\left(\mathbf{1}\right)\mathbf{=}\mathbf{P}\left(\mathbf{2}\right)\mathbf{=}\mathbf{P}\left(\mathbf{3}\right)\mathbf{=}\mathbf{P}\left(\mathbf{4}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{16}}$and $\mathbf{P}\left(\mathbf{5}\right)\mathbf{=}\mathbf{P}\left(\mathbf{6}\right)\mathbf{=}\mathbf{P}\left(\mathbf{7}\right)\mathbf{=}\mathbf{P}\left(\mathbf{8}\right)\mathbf{=}\mathbf{P}\left(\mathbf{9}\right)\mathbf{=}\mathbf{P}\left(\mathbf{10}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}$ . Find $\mathbf{P}\left(\mathbf{A}\right)$and $\mathbf{P}\left(\mathbf{B}\right)$.

Short answer

1. Venn Diagram
2. $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}$and$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{10}}$
3. $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{5}}{\mathbf{16}}$and$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Step-by-step solution

Explaining the type of diagram given in the question

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.

Finding  PA and  PB

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 $\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{10}}$.

Finding PA  and src="data:image/svg+xml;base64,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" role="math" localid="1649073820808" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/7b836d52-4c59-4ba4-b3b1-3ae96c45bcc1.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220530%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220530T022715Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=c769ce3daa91036c8c0556bd6afd84739db874ba9dc92082a32da59326ce5fe6" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/7b836d52-4c59-4ba4-b3b1-3ae96c45bcc1.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220530%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220530T022446Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=d2dbe1f1915dec897529041a53a418744c12408130616aa34aac7ebd477d9a38" PB with the help of given equation

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$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$.