Solutions of Question 11 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Given the following spinner, determine the probability.
Total number colors $$n(S)=4$$ P(orange) The orange color covers one fourth $\dfrac{1}{4}$ of the spinner,
thus the probability of is: $$\quad P( orange )=\dfrac{1}{4}$$
P(Red or Green)
Red color cover one fourth $\dfrac{1}{4}$ and green color ccvers one fourth $\dfrac{1}{4}$ of the spinner.
Therefore, \begin{align}P(\operatorname{Red})&=\dfrac{1}{4}\\ P( Green )&=\dfrac{1}{4}\end{align} Also these two are mutually exclusive events.
Therefore $P(R \cap G)=\phi$, where $R$ stands for red event and $G$ stands for green event.
By addition law of probability, we have \begin{align}\boldsymbol{P}( Red or Green )&=P(\text { Red })+P( Green )-P( Red and Green )\\ \Rightarrow P(Red or Green )&=\dfrac{1}{4}+\dfrac{1}{4}-\phi\\ &=\dfrac{1}{2}\end{align} $\mathbf{P}( Not Red)$ The probability of red is: $$P(\text { Red })=\dfrac{1}{4}$$ Then by complementary event theorem: \begin{align} P(\text { not red })&=1-P(\text { Red }) \\ & =1-\dfrac{1}{4}=\dfrac{3}{4}\end{align}
$P(Pink)$
Since pink color covers one fourth $\dfrac{1}{4}$ of the spinner,
thus the probability of is: $$P( pink )=\dfrac{1}{4}$$ $$\text{Hence} \dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4},\dfrac{1}{4}$$