Hello I am having some trouble understanding why my thinking is not correct and I am having trouble understanding the solution.

The question begins by saying suppose we have a fair coin that we will flip twice and let the events be denoted A= Head on toss 1 B= Head on toss 2 and C= same side both toss

So I listed $$S=\{(H,H),(H,T),(T,H),(T,T)\}$$

$$A=\{(H,H),(H,T)\}$$ $$B=\{(H,H),(T,H)\}$$ $$C=\{(H,H),(T,T)\}$$

with $P[A]=P[B]=P[C]=\frac{1}{2}$ Then it asks some questions about independence which I am fine with, as I know that A and B are independent if $P[A \cap B]=P[A]P[B]$

Then it asks to show that $(A \cup B)$ and C are independent, so what I thought was

$A \cup B =\{(H,H),(H,T),(T,H)\}$ with $P[A \cup B]= \frac{3}{4}$

and $(A \cup B) \cap C= \{(H,H)\}$ $P[(A \cup B) \cap C]=\frac{1}{4}$

But $$\frac{3}{4}\frac{1}{2}=\frac{3}{8} \neq \frac{1}{4}$$ so it seemingly says they are not independent.

So where am I going wrong, where is my mistake? Thanks

Actually the mistake was in my understanding of quesiton, I have now solved it.

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    $\begingroup$ Are you sure you're supposed to show they're independent? Your analysis looks right to me; $A \cup B$ and $C$ are not independent. $\endgroup$ – Brian Tung Sep 24 '15 at 16:38
  • $\begingroup$ Yes it is question 2.176 In Wackerlys mathematical statistics with applications, seventh edition $\endgroup$ – Quality Sep 24 '15 at 16:39
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    $\begingroup$ They are not independent as you have shown. Intuitively, look at $P(C | A \cup B) = {1 \over 3} \neq P(C) = {1 \over 2}$. $\endgroup$ – copper.hat Sep 24 '15 at 16:44

You misunderstood the exercise. According to this copy of the $7$th edition, Exercise $2.176$ does not ask you to show that $A\cup B$ and $C$ are independent for these events $A$, $B$ and $C$, but for mutually independent events $A$, $B$ and $C$. It refers back to Exercise $2.175$ for the definition of mutually independent events. It does not refer back to that exercise for this particular example of $A$, $B$, and $C$, as the correct answer to Exercise $2.175$ is that these events are not mutually independent.

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  • $\begingroup$ Thanks then, I think it could maybe be a bit more clear with that but I understand now $\endgroup$ – Quality Sep 24 '15 at 18:19

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