Question

Consider the Venn diagram below, were

$\begin{array}{l}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{3}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{,}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{5}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{20}}\\ \mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{6}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{,}\mathbf{andP}\mathbf{\left(}{\mathbf{E}}_{\mathbf{7}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\end{array}$

Find each of the following probabilities:

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\\ \mathbf{b}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\\ \mathbf{c}\mathbf{.}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\\ \mathbf{d}\mathbf{.}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\end{array}$ $\begin{array}{l}\mathbf{e}\mathbf{.}\mathbf{P}\mathbf{\left(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\right)}\\ \mathbf{f}\mathbf{.}\mathbf{P}\mathbf{\left(}{\mathbf{B}}^{\mathbf{c}}\mathbf{\right)}\\ \mathbf{g}\mathbf{.}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }{\mathbf{A}}^{\mathbf{c}}\mathbf{)}\\ \mathbf{h}\mathbf{.}\mathbf{P}\mathbf{(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\cap }\mathbf{B}\mathbf{)}\end{array}$

Short answer

1. 3/4
2. 7/10
3. 1
4. 2/5
5. 1/4
6. 7/20
7. 1
8. 3/10

Step-by-step solution

By considering the Venn diagram, find the probability

A Venn diagram is a probability diagram with one or more circles inside a rectangle and demonstrates logical relationships between occurrences. In a Venn diagram, the rectangle symbolizes the sample space or the universal set, which is the collection of all possible outcomes.

We know that probability $\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{}\sum _{i=1}^x{x}_i$

Were,

localid="1653540716903" ${\mathbf{x}}_{\mathbf{i}}$are the events belonging to x.

So,

localid="1662214298829" $$\begin{gathered} \mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{5}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{6}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{2}}{\mathbf{20}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{15}}{\mathbf{20}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}} \end{gathered}$$

Find the probability of P (B)

$$\begin{gathered} \mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{4}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{7}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{4}}{\mathbf{20}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{7}}{\mathbf{10}} \end{gathered}$$

Find the probability of P(A∪B)

$$\begin{gathered} \mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{4}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{5}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{6}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{7}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{4}}{\mathbf{20}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{20}}{\mathbf{20}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1} \end{gathered}$$

Find the probability

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{3}}\mathbf{\right)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{1}}{\mathbf{5}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}\end{array}$

Find the probability

$\begin{array}{c}\mathbf{P}\mathbf{\left(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{4}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{7}}\mathbf{\right)}\\ \mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\\ \mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{4}}{\mathbf{20}}\\ \mathbf{=}\frac{\mathbf{5}}{\mathbf{20}}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\end{array}$

Find the probability

$\begin{array}{l}\mathbf{P}\mathbf{\left(}{\mathbf{B}}^{\mathbf{c}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{5}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{E}}_{\mathbf{5}}\mathbf{\right)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{10}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{2}}{\mathbf{20}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{7}}{\mathbf{20}}\end{array}$

Find the probability

$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }{\mathbf{A}}^{\mathbf{c}}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }{\mathbf{A}}^{\mathbf{c}}\mathbf{)}$

Here,

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }{\mathbf{A}}^{\mathbf{c}}\mathbf{)}\mathbf{=}\mathbf{0}\\ \mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\\ \mathbf{P}\mathbf{\left(}{\mathbf{A}}^{\mathbf{C}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\end{array}$

Hence,

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }{\mathbf{A}}^{\mathbf{c}}\mathbf{)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{-}\mathbf{0}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{+}\mathbf{1}}{\mathbf{4}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}}{\mathbf{4}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\end{array}$

Find the probability

$\begin{array}{l}\mathbf{P}\mathbf{(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{7}}{\mathbf{10}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{5}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{7}\mathbf{-}\mathbf{4}}{\mathbf{10}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}\end{array}$