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)=(n(E_1))/(n(S))=(9)/(36)=(1)/(4)`
`P(E_2)=(n(E_2))/(n(S))=(6)/(36)=(1)/(6)`
`P(E_1\cap E_2)=(n(E_1\cap E_2))/(n(S))=(3)/(36)=(1)/(12)`

Therefore,

`P(E_1)+P(E_2)+P(E_1\cap E_2)=(1)/(4)+(1)/(6)+(1)/(12)=(3+2+1)/(12)=(6)/(12)=(1)/(2)`

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