====== Question 11 Review Exercise 6 ======
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.

=====Question 11=====
Given the following spinner, determine the probability.
====Solution====
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: $$P( pink )=\dfrac{1}{4}$$
$$\text{Hence} \dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4},\dfrac{1}{4}$$


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