Answer:
The total number of sections on the spinner indicates the sample space
[tex]\begin{gathered} n\left(S\right)=\left\lbrace1,2,3,4,5,7\right\rbrace \\ n\left(S\right)=6 \end{gathered}[/tex]
Let the event of the spinner showing an even number be
[tex]E_1[/tex][tex]\begin{gathered} E_1=\left\lbrace2,4\right\rbrace \\ n\left(E_1\right)=2 \end{gathered}[/tex]
To figure out the probability of getting an even number on the spinner, we will use the formula below
[tex]Pr\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} Pr(E_{1})=\frac{n(E_{1})}{n(S)} \\ Pr(E_1)=\frac{2}{6} \\ Pr(E_1)=\frac{1}{3} \end{gathered}[/tex]
Hence,
The final answer is 1/3
OPTION D is the right answer
