Why is this reasoning about probability invalid? I have a confusion about probability. I know my reasoning is wrong but I can't figure out why it's wrong. Can someone please clarify. 
If I flip a coin once, the probability that it lands heads is $\frac{1}{2}$. If I flip it again, the probability is again $\frac{1}{2}$. These are independent events so I can find out the probability that I either the first one or the second one is heads by adding them up. So, I get $\frac{1}{2} + \frac{1}{2} = 1$. (reductio ad absurdum:) This can be extended to three flips where probability becomes $\frac{3}{2}$. 
p.s. I know we need to do $\{HH, HT, TH, TT\}$ for correct answer but what's wrong with the reasoning above?  
 A: It is always wrong to add probabilities of independent events. Adding probabilities is for mutually exclusive events. But independent events are not mutually exclusive -- "independent" means (speaking figuratively and sloppily) that one events doesn't know whether the other has happened or not, so it is perfectly possible for both of them to happen.
A: This is the inclusion-exclusion principle - you're including the case that both the first and second flip give heads twice. To see this more clearly, you can look at the four outcomes HH, HT, TH and TT that can occur. The probability that the first coin comes up heads is the probability that you get HH or HT, and the probability that the second comes up heads is the probability that you get HH or TH.
So when you add them together, you've counted the probability of getting HH twice. So you need to subtract this probability, which is $\frac{1}{4}$, to get the correct answer of $\frac{3}{4}$ that either the first or the second coin toss is heads.
In general, for independent events $A$ and $B$:
$$P(A\vee B)=P(A)+P(B)-P(A\wedge B)$$
