Question

1) find the probability the arrow will land on a multiple of 4 or 6 or both

Answer 1

Write out the number of elements in the sample space as we denote the sample space as S

$n(S)=36$

Write out the event space E1 for the multiples of 4

$\begin{gathered} E_1=\mleft\lbrace4,8,12,16,20,24,28,32,36\mright\rbrace \\ n(E_1)=9 \end{gathered}$

Write out the event space E2 for the multiples of 6

$\begin{gathered} E_2=\mleft\lbrace6,12,18,24,30,36\mright\rbrace \\ n(S)=6 \end{gathered}$

The probability that the arrow will land on a multiple of 4 or 6 or both will land can be represented thus

$P(E_1)+P(E_2)+P(E_1\cap E_2)$$\begin{gathered} E_1\cap E_2=\mleft\lbrace12,24,36\mright\rbrace \\ n(E_1\cap E_2)=3 \end{gathered}$$P(E_1)=\frac{n(E_1)}{n(S)}=\frac{9}{36}=\frac{1}{4}$$P(E_2)=\frac{n(E_2)}{n(S)}=\frac{6}{36}=\frac{1}{6}$$P(E_1\cap E_2)=\frac{n(E_1\cap E_2)}{n(S)}=\frac{3}{36}=\frac{1}{12}$

Therefore,

$P(E_1)+P(E_2)+P(E_1\cap E_2)=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3+2+1}{12}=\frac{6}{12}=\frac{1}{2}$

Hence, the probability that the arrow will land on a multiple of 4or 6 or both is 1/2