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Given that two independent tosses of a fair coin. Let $A$be the event that the first toss results in heads, let $B$be the event that the second toss results in heads, and let $C$ be the event that in both tosses the coin lands on the same side.

On the off chance that a fair coin is thrown twice freely there are 4 similarly reasonable events:

$(H,H)\text{- both tosses resulted in heads,}⇒P(H,H)=P\left(H\right)â‹\dots P\left(H\right)=\frac{1}{2}â‹\dots \frac{1}{2}=\frac{1}{4}$

$(H,T)\text{- first toss is heads, the second tails,}P(H,T)=\frac{1}{4}$

$(H,T)\text{- first toss is tails, the second heads}P(T,H)=\frac{1}{4}$

$(T,T)\text{- both tosses resulted in tails,}P(T,T)=\frac{1}{4}$

These events are all mutually exclusive.

Defined events:

$P\left(A\right)=P\left(\right\{(H,H),(H,T)\left\}\right)=P\left(\right\{(H,H)\left\}\right)+P\left(\right\{(H,T)\left\}\right)=2\frac{1}{4}=\frac{1}{2}$

$P\left(B\right)=P\left(\right\{(H,H),(T,H)\left\}\right)=\frac{1}{2}$

$P\left(C\right)=P\left(\right\{(H,H),(T,T)\left\}\right)=\frac{1}{2}$

Characterization of independence is that the probability of intersection is the result of the probabilities:

$P\left(AB\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}â‹\dots \frac{1}{2}=P\left(A\right)P\left(B\right)⇒A\text{and}B\text{are independent}$

$P\left(BC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}â‹\dots \frac{1}{2}=P\left(B\right)P\left(C\right)⇒B\text{and}C\text{are independent}$

$P\left(AC\right)=P\left(\right\{(H,H)\left\}\right)=\frac{1}{4}=\frac{1}{2}â‹\dots \frac{1}{2}=P\left(A\right)P\left(C\right)⇒A\text{and}C\text{are independent}$

All events are independent in pairs.

Thus all three events are not independent: