Two events are independent if
[tex]P(A)\cdot P(B)=P(B\cap A)[/tex]
Or
[tex]\begin{gathered} P(B)=\frac{P(B\cap A)}{P(A)} \\ P(B)=P(B|A) \end{gathered}[/tex]
Event A is "rolling a 5":
[tex]P(A)=\frac{1}{6}[/tex][tex]P(B\cap A)=\frac{1}{6}[/tex]
Event B is "rolling an odd number":
Odd number are 1, 3 and 5
[tex]P(B)=\frac{3}{6}=\frac{1}{2}[/tex]
Then, substitute
[tex]\begin{gathered} P(B)=\frac{P(B\cap A)}{P(A)} \\ P(A)=\frac{P(B\cap A)}{P(B)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3} \end{gathered}[/tex]
This is:
[tex]\begin{gathered} P(B)\ne\frac{P(B\cap A)}{P(A)} \\ therefore \\ P(B)\ne P(B|A) \end{gathered}[/tex]
which makes the events dependent.
Answer: D.
