# Proof: Why Probability of complement of A equals to one minus Probability of A [ P(A’) = 1-P(A) ]

**Probability** refers to the extent of the occurrence of events. When an event occurs like throwing a ball, picking a card from the deck, etc., then there must be some probability associated with that event.

**Basic Terminologies:**

**Random Event-**

If the repetition of an experiment occurs several times under similar conditions, if it does not produce the same outcome every time, but the outcome in a trial is one of the several possible outcomes, then such an experiment is called a random event or a probabilistic event.**Sample Space****–**

Sample Space refers to the set of all possible outcomes of a random event. For example, when a coin is tossed, the possible outcomes are head and tail.**Event****–**

An event refers to the subset of the sample space associated with a random event.**Occurrence of an Event****–**

An event associated with a random event is said to occur if any one of the elementary events belonging to it is an outcome.**Complement****–**

In sets, the complement of a set, A is the set of all elements which are not present in A. It is denoted by A’ or A^{c}. For example, in an experiment of rolling a die, if A is the set of all even outcomes, then, A’ is the set of all odd outcomes.**Mutually Exclusive Events****–**

Two or more events associated with a random event are said to be mutually exclusive events. If any one of the events occurs, it prevents the occurrence of all other events. This means that no two or more events can occur simultaneously at the same time.

If A and B are mutually exclusive events, then A ∩ B = ∅ Also, P(A ∩ B) = 0

**Probability Axioms:**

- For any set A, the probability of A will always be greater than or equal to zero, i.e. P(A) >=0
- The probability of Sample Space(S) will always be equal to one i.e. P(S) = 1.
- If A
_{1}, A_{2}, A_{3}, A_{4}… A_{N}are mutually exclusive events, then the probability of the union of these mutually exclusive events will be equal to the sum of the probability of these mutually exclusive events, i.e. P(A_{1}∪ A_{2}∪ A_{3}∪ A_{4}∪ …. A_{N}) = P(A_{1}) + P(A_{2}) + P(A_{3}) + P(A_{4}) + …. + P(A_{N})

**Problem Statement:**

Why does the probability of the complement of A equal one minus the probability of A?

**Solution:**

Let’s get started with the proof of the above problem statement. Below are the steps for the proof-

1. Consider an event, A. Since the sample space of an experiment contains all the possible outcomes of an experiment and the union of A and A’ comprises all the possible outcomes of the experiment. So, sample space can be written as-

S = A ∪ A' Also, P(S) = P(A ∪ A') --- (1)

2. Since the intersection of A and A’ equals ∅, it can be said that A and A’ are mutually exclusive events. So, according to axiom 3,

Since, A ∩ A' = ∅ P(A ∪ A') = P(A) + P(A') --- (2)

3. Now, from equation (1) and (2), it can be written as-

P(S) = P(A) + P(A')

4. From axiom 2, it is known that the probability of a sample space always equals 1 i.e.

P(S) = 1 P(A) + P(A') = 1 --- (3)

5. After rearranging the equation (3), the following equation is obtained-

P(A') = 1 - P(A)

i.e. the probability of the complement of A equals one minus the probability of A. Hence, Proved.

**Examples:**

Let’s have a look at some solved examples related to the above proof-

**1. A die is thrown once.**

Event A-An even number appears

Sample Space, S-{1, 2, 3, 4, 5, 6}P(S) = 1

A = {2, 4, 6} i.e. P(A) = 3/6 = 1/2

A’ = {1, 3, 5} i.e. P(A’) =3/6 = 1/2

Now, we can easily observe that S = A ∪ A’ and since A ∩ A’ = ∅,

A and A’ are mutually exclusive events, which impliesP(S) = P(A ∪ A’)

P(S) = P(A) + P(A’)

P(S) = 1/2 + 1/2

= 1

P(A’) = 1 – P(A)

P(A’) = 1 – 1/2

P(A’) = 1/2

**2. ****Two coins are tossed.**

Event A-Two head appears

Sample Space, S- {HH, HT, TH, TT}P(S) = 1

A = {HH} i.e. P(A) =1/4

A’ = {HT, TH, TT} i.e. P(A’) =3/4

i.e. S = A ∪ A’ and since A ∩ A’ = ∅, we can say that A and A’ are

mutually exclusive events, which implies that –P(S) = P(A ∪ A’)

P(S) = P(A) + P(A’)

P(S) = 1/4 + 3/4

P(S) = 1

P(A’) = 1 – P(A)

P(A’) = 1 – 1/4

P(A’) = 3/4

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