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) = mA/m where,
- P(A) is the probability of A
- mA 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.
There are so many errors in two of the lectures that I have watched. The flow of the lectures are also inappropriate. Firstly you never defined what an event is. For this lecture you can just say that an event is a subset of sample space. Therefore it can be any subset of sample space, even phi(empty set) or the whole sample space itself. You are confusing events with elements of sample space.
There is a fundamental errors on tis page too. Like P(A|B) is probability of event A given that event B has (already) occurred. However the text in this page says the other way.
Please revise it before making public.
Your well-wisher
Kumar.
@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events.
Ex. Tossing a coin and rolling a die.
Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.”
The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample space is S as given but the even space is E= {(1,H),(1,T)} this is a compound event as the event has 2 possible outcomes that fit and not just one. ( and also they are independent.)
@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events.
Ex. Tossing a coin and rolling a die.
Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.”
The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample space is S as given but the even space is E= {(1,H),(1,T)} this is a compound event as the event has 2 possible outcomes that fit and not just one. ( and also they are independent.)