Question

The sample space for an experiment contains five sample points with probabilities as shown in the table. Find the probability of each of the following events:

a. Either 1,2 or 3 occurs

b. Either 1,3 or 5 occurs

c. 4 does not occur

Short answer

$\mathbf{a}\mathbf{.}\mathbf{}\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{55}\mathbf{,}\mathbf{}\mathbf{b}\mathbf{.}\mathbf{}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{,}\mathbf{}\mathbf{c}\mathbf{.}\mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{C}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}$

Step-by-step solution

Finding the probability of event  A

To calculate the probability of event A happening, we add the individual probabilities of sample points 1, 2 and 3.

Given,

A: Either 1,2 or 3 occurs

$\begin{array}{l}\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{P}\left(\mathbf{1}\right)\mathbf{+}\mathbf{P}\left(\mathbf{2}\right)\mathbf{+}\mathbf{P}\left(\mathbf{3}\right)\\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{30}\\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{55}\end{array}$$$\begin{gathered} \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{P}\left(\mathbf{1}\right)\mathbf{+}\mathbf{P}\left(\mathbf{2}\right)\mathbf{+}\mathbf{P}\left(\mathbf{3}\right) \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{30} \\ \mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{55} \end{gathered}$$

Hence, $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{55}$$\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{55}$

Finding the probability of event  B

To calculate the probability of event B happening, we add the individual probabilities of sample points 1, 3 and 5.

B: Either 1,3 or 5 occurs

$\begin{array}{l}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{P}\left(\mathbf{1}\right)\mathbf{+}\mathbf{P}\left(\mathbf{3}\right)\mathbf{+}\mathbf{P}\left(\mathbf{5}\right)\\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{15}\\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\end{array}$$$\begin{gathered} \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{P}\left(\mathbf{1}\right)\mathbf{+}\mathbf{P}\left(\mathbf{3}\right)\mathbf{+}\mathbf{P}\left(\mathbf{5}\right) \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{15} \\ \mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50} \end{gathered}$$

Hence, $\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$

Finding the probability of event C

To calculate the probability of event C happening, we subtract the probability of sample point 4 occurring from 1.

C: 4 does not occur

$\begin{array}{l}\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{P}\left(\mathbf{C}\right)\\ \mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{30}\\ \mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}\end{array}$$$\begin{gathered} \mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{P}\left(\mathbf{C}\right) \\ \mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{30} \\ \mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70} \end{gathered}$$

Hence,$\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}$$\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}$