Question

An experiment consists of rolling a standard six-sided die once.

Answer 1

Two events are independent if

`P(A)\cdot P(B)=P(B\cap A)`

Or

`\begin{gathered} P(B)=(P(B\cap A))/(P(A)) \\ P(B)=P(B|A) \end{gathered}`

Event A is "rolling a 5":

`P(A)=(1)/(6)`
`P(B\cap A)=(1)/(6)`

Event B is "rolling an odd number":

Odd number are 1, 3 and 5

`P(B)=(3)/(6)=(1)/(2)`

Then, substitute

`\begin{gathered} P(B)=(P(B\cap A))/(P(A)) \\ P(A)=(P(B\cap A))/(P(B))=((1)/(6))/((1)/(2))=(1)/(3) \end{gathered}`

This is:

`\begin{gathered} P(B)\\e(P(B\cap A))/(P(A)) \\ therefore \\ P(B)\\e P(B|A) \end{gathered}`

which makes the events dependent.

Answer: D.