# 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?

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The [logic] tag is not for questions about "logical reasoning". It is for questions about mathematical logic. – Asaf Karagila Dec 29 '12 at 12:01
Apologies. Thanks for the edit. – recluze Dec 29 '12 at 13:18

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 $B$:

$$P(A\vee B)=P(A)+P(B)-P(A\wedge B)$$

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This one is a very clear answer and is good for explaining stuff to beginners. Thanks . – recluze Dec 29 '12 at 13:17

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.

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Indeed, two independent events are mutually exclusive if and only if at least one of them has probability $0$. – Zhen Lin Dec 29 '12 at 12:06
@ZhenLin: And perhaps not even then. Suppose $X$ and $Y$ are independent normal distributed variables. Then $X=0$ and $Y=0$ are independent and both have probability 0, but they are not mutually exclusive. – Henning Makholm Dec 29 '12 at 12:11
Well, almost never is as good as never when it comes to computing probability, so... – Zhen Lin Dec 29 '12 at 12:14
It is not always wrong. What is wrong is to expect the sum to express joint probability, as opposed to expected value of the number of events that do happen. Unless, of course, at most one of the events has nonzero probability. :) (Of course, for that, independence isn't needed.) – tomasz Dec 29 '12 at 13:10
@tomasz: But then what you're adding are not probabilities, but expectations (which happen to be numerically equal to some probabilities). Expectations can be added without any assumptions about independence. – Henning Makholm Dec 29 '12 at 13:13