Question

An experiment results in one of the following sample points: ${\mathbf{E}}_{\mathbf{1}}\mathbf{,}{\mathbf{E}}_{\mathbf{2}}\mathbf{,}{\mathbf{E}}_{\mathbf{3}}$ or${\mathbf{E}}_{\mathbf{4}}$ . Find $\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)$for each of the following cases.

1. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$
2. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)$
3. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$and$\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)$

Short answer

1. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$
2. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$
3. $\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$

Step-by-step solution

Finding the Probability of the event E4  where  PE1=0.1,PE2=0.2,PE3=0.3

The probability of all the sample points within a sample space is equal to $\sum \mathbf{Pi}\mathbf{=}\mathbf{1}$

Given $\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$

$$\begin{gathered} \mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1}\mathbf{-}\left(\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3}\right) \\ \mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4} \end{gathered}$$

The probability of observing the event ${\mathbf{E}}_{\mathbf{4}}$ is 0.4.

Finding the probability of the event E4  where PE1 = PE2 = PE3 = PE4

The probability of all the sample points within a sample space is equal to $\sum \mathbf{Pi}\mathbf{=}\mathbf{1}$.

Given,

$\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)$

$$\begin{gathered} \mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{4}\mathbf{\times }\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}} \\ \mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25} \end{gathered}$$

The probability of observing event ${\mathbf{E}}_{\mathbf{4}}$ is$0.25$ .

Finding the probability of the event E4  where PE1=PE2=0.1  and  PE3=PE4

The probability of all the sample points within a sample space is equal to $\sum \mathbf{Pi}\mathbf{=}\mathbf{1}$.

Given,

$\mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$and $\mathbf{P}\left({\mathbf{E}}_{\mathbf{3}}\right)\mathbf{=}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)$

$$\begin{gathered} \mathbf{P}\left({\mathbf{E}}_{\mathbf{1}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{2}}\right)\mathbf{+}\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1} \\ \mathbf{2}\mathbf{\times }\mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{1}\mathbf{-}\left(\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{1}\right) \\ \mathbf{P}\left({\mathbf{E}}_{\mathbf{4}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4} \end{gathered}$$

The probability of observing the event ${\mathbf{E}}_{\mathbf{4}}$ is 0.4 .