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What I seen so far,

The probability of tossing heads is 1/2.
The probability of tossing tails is 1/2.

Therefore, the probability of tossing a coin for either tails or heads is 1 which is a mutually exclusive event.

Is it possible to show an example of non-mutually exclusive event using such a single coin example? If not then what are it's requirements?

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What is your understanding of the phrase "mutually exclusive event"? A single event cannot be mutually exclusive; you need at least two events for mutual exclusion to occur. Also, if the probabilities of two events sum to $1$, that does not imply that the events must be mutually exclusive. –  Dilip Sarwate Oct 11 '12 at 2:59
In my example, the first event is tossing heads and the second event is tossing tails. So, they are mutually exclusive i.e. having no outcomes in common and the probability for either of event is 1. –  user121314 Oct 11 '12 at 3:21
Yes, but you say :...tossing a coin for either heads or tails... which is a mutually exclusive event." (emphasis added). The single event "either heads or tails" which you correctly say has probability $1$ cannot be a mutually exclusive event; as I said in my first comment, mutual exclusion requires at least two events. The two events "outcome is head" and "outcome is tail" are mutually exclusive events: the single event "either heads or tails" cannot be called a mutually exclusive event. –  Dilip Sarwate Oct 11 '12 at 13:17

1 Answer 1

up vote 1 down vote accepted

Event A: The coin comes up heads.
Event B: The coin comes up either heads or tails.

The probability of A is $\frac12$; the probability of B is $1$.

For less trivial examples you need more than two possible outcomes. Flipping a coin twice (or flipping two coins) will work, as will rolling a die. With two coin tosses, for instance:

Event A: I get at least one head.
Event B: I get at least one tail.

These are not mutually exclusive: both occur if I get HT or TH. They’re also not identical: if I get HH, A occurs but B doesn’t. Each has probability $\frac34$.

With a die:

Event A: the number that comes up is even.
Event B: The number that comes up is $1,2$, or $3$.

If I roll a $2$, both A and B occur, so they’re not mutually exclusive. Note that in this case each has probability $\frac12$, so their probabilities do add up to $1$, even though they are not mutually exclusive.

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To add a little to dispel what seems to be another idea embedded in the OP's mind, the events $A$ and the event $C$: "roll a $3$" are mutually exclusive events of probabilities $\frac{1}{2}$ and $\frac{1}{6}$. Thus, $P(A\cup C) = \frac{2}{3} < 1$ showing that it is not necessary that the union of mutually exclusive events have probability $1$; smaller probabilities are possible. It is true, though, that the probability of the union of mutually exclusive events cannot exceed $1$. –  Dilip Sarwate Oct 11 '12 at 13:25

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