Out of a collection of events, if at a given time the occurrence of only one of them is possible. Then, that collection is a collection of mutually exclusive events. For example, the event of a person being an adult. The person is either an adult or not an adult, no in between. So, these two events are mutually exclusive.

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## Mutually Exclusive Events

If the occurrence of an event makes the occurrence of another event impossible for that particular experiment then they are mutually exclusive. It becomes an ‘either-or’ situation.

- For example, the outcome after rolling a die can be
*either*an even number*or*an odd number. - So, {1, 3, 5} and {2, 4, 6} are sets of mutually exclusive events.

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## Mutually Exclusive and Exhaustive Events

The condition for mutually exclusive events for being exhaustive is the outcome of an experiment must be one out of the sample space of mutually exclusive events for that particular experiment.

For example, the blood group of a person. The events are {A, B, AB, O}. These are the only possible blood groups and a person can have only one of these. So, we can call blood test as an experiment and all the possible outcomes as mutually exclusive and exhaustive events. *These events are equally likely as well.*

### Definition

In case of the events being mutually exclusive, exhaustive and equally likey the probability is

P(A) = m_{A}/m where,

- P(A) is the probability of A
- m
_{A}is the no. mutually exclusive, exhaustive and equally likely events favourable to event A. - m is the total no. of mutually exclusive, exhaustive and equally likely events.

*The *above-given* definition is *the classical* definition of probability. It is applicable *only* when the events are finite and equally likely. *Taking set theory into consideration we can write the definition of probability as:

Here,

- S = sample space of finite no. elementary events for an experiment.
- A = A ⊂ S, A is an event under consideration then,
- P(A) = n(A)/n(S) where,
- n(S) denotes the total no. of events in S.
- n(A) denotes the no. of outcomes favourable to A.

Now, complement of A(i. e., A’) will be: n(A’) = n(S) – n(A).

⇒n(A’)+ n(A) = n(S)

A and A’ make a pair of mutually exclusive and exhaustive events. This gives, for any two mutually exclusive events say A and B.

- A ∩ B = Φ ⇒ n(A ∩ B) = 0
- ⇒P(A ∩ B) = 0; 0 is the minimum possible which is the probability of an impossible event.
- (A ∩ B) is an impossible event because they are exclusive and t is impossible for them to occur simultaneously.
- P(A ∪ B) = P(A) + P(B)

For mutually exclusive and exhaustive events P(A) + P(B) = 1 because

- The maximum possible probability is 1, which is the probability of a sure event.
- Since P(A) and P(B) are exhaustive they are the only two events.
- ∴ P(A ∪ B) = 1; (A ∪ B) is a sure event as one of the two events are sure to occur for the experiment.

## Solved Examples on Mutually Exclusive Events

Question:** **An urn contains balls of various colours. The probability of drawing red is 1/3, blue is 1/7, green is 1/9. If a ball is drawn at random what is the probability that

- Either red or blue is drawn
- Neither red nor blue is drawn
- None out of red, blue, green gets drawn

Solution. a) It is given that, P(R) = 1/3 and P(B) = 1/7

∴ P(R∪B) = P(R) + P(B)

= 1/3 + 1/7 = 10/21*All three events R, B and G are exclusive and that’s why their individual probabilities are added*

b) P(R∪B)’ = 1 – P(R∪B)

= 1 – 10/21 = 11/21*Notice here that since the events R, B, and G are exclusive but not exhaustive P(R∪B)≠P(G).*

c) P(R∪B∪G)’ = 1 – P(R∪B∪G)

= 1 – (1/3 + 1/7 + 1/9)

= 1 – 29/45 = 16/45*As the sets are not exhaustive P(R∪B∪G)’≠0.*