Showing Conditional Independence Does Not Imply Independence

[Jason4]

I know this isn't quite advanced probability, but I'm not sure if I have this right.

I want to show that conditional independence of $X$ and $Y$ given $Z$ does not imply independence of $X$ and $Y$ (and vice versa).

So I used coin tosses where:

$X=\{$ first coin tails $\}$

$Y=\{$ second coin tails $\}$

$Z=\{$ both coins same $\}$

I can show that independence does not imply conditional independence.

How do I show that conditional independence does not imply independence?

 

[fawaz1]

Well to answer such a question, you need to take two events that are not independent and show they are conditionally independent given a third event. Example:A basketball player has two shots.Let A be the event that the player scores the first shot. Assume P(A) = 0.3Let B be the event that the player scores the second shot. Assume P(B/A) = 0.2 and P(B/A' ) = 0.4 (if he/she scores the first shot, he/she has less probability of scoring the second)Let C be the event that both shot are scored. Clearly, A and B are not independent. Try to find P(A/C) and P(B/C) and P( (A and B)/C) and prove that P(A/C)*P(B/C) = P( (A and B) /C).