The addition theorem of probability is needed as noting down all the elements of a sample space and counting the total number of favorable outcomes can be a lengthy process. Probability is the possibility or chance of occurring an event. Probability of occurrence of an event A, i.e. P(A), closely associated with the discrete sample space, is the sum of probabilities assigned to the sample points in A.

What is Addition Theorem of Probability?

Probability is the chances of occurrence of event A in a given sample space C. With the help of the addition theorem of probability, when multiple events are given, the probability of occurring of one of the events can be easily computed.

Maths Notes Free PDFs

Symbols used in probability:

\(A\cup B\): This represents the occurrence of at least one event A or B. This means either A occurs or B occurs or both occur simultaneously.

\(A\cap B\): This represents the occurrence of both the events A and B simultaneously.

\(\overline{A}\): A does not occur.

\(\overline{A}\cap\overline{B}\): Neither A nor B occurs.

\(\overline{A}\cap B\): A does not occur but B occurs.

\(\left(A\cap\overline{B}\right)\cup\left(\overline{A}\cap B\right)\): Exactly one of the two events A or B occurs.

Let us now understand the theorem and proof for addition theorem of probability in detail.

Two events are mutually exclusive if they cannot occur together in a single experiment. If one event happens, the other cannot.

Example:
When you flip a coin, you can either get Heads or Tails, but not both at once. So, "Getting Heads" and "Getting Tails" are mutually exclusive events.

Addition Theorem of Probability for Mutually Exclusive Events

Addition Theorem of Probability for Mutually Exclusive Events Statement: If A and B are two mutually exclusive events then the probability of occurrence of A or B is the sum of individual probabilities of A and B.

This can be represented as:

\(P\left(A\cup B\right)=P\left(A\ or\ B\right)=P\left(A\right)+P\left(B\right)\)

Addition Theorem of Probability for Mutually Exclusive Events Proof:

Let N be the total number of exhaustive and equally likely cases of an experiment.

Let \(m_1 and m_2\) be number of cases favorable to happening of events A and B respectively.

Then,\(P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{m_1}{N}\)and \(P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{m_2}{N}\).

Since A and B are Mutually exclusive events, total number of events favorable to either A or B i.e. \(n\left(A\cup B\right)=m_1+m_2\), then,

\(P\left(A\cupB\right)=\frac{n\left(A\cupB\right)}{n\left(S\right)}=\frac{n\left(A\cup B\right)}{N}=\frac{m_1+m_2}{N}=\frac{m_1}{N}+\frac{m_2}{N}=P\left(A\right)+P\left(B\right)\).

In general:

If we extend the above theorem to 2 or more mutually exclusive events.

The probability of occurrences of any one of the several mutually exclusive events A,B, and C is equal to sum of their individual probabilities.

This can be given as:

\(P\left(A\cupB\cupC\right)=P\left(A\or\B\or\C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)\).

Generalizing the result:

If \(A_1,A_2,…,A_n\) are mutually exclusive events, then,

\(P\left(A_1\cup A_2\cup…\cup A_n\right)=P\left(A_1\right)+P\left(A_2\right)+…+P\left(A_n\right)\).

i.e. the probability of occurrence of any one of ‘n’ mutually exclusive events \(A_1, A_2, …, A_n\) is equal to the sum of individual probabilities.

Learn about the difference between mutually exclusive and independent events here.

Non-Mutually Exclusive Events :

Two events are called non-mutually exclusive if they can happen at the same time in one experiment. For example, when rolling a die, getting an odd number (A = {1, 3, 5}) and getting a 3 (B = {3}) can both happen together because 3 is part of both events.

Addition Theorem of Probability for Non-Mutually Exclusive Events

The theorem of probability discussed above is not applicable to non-mutually exclusive events.

Let us discuss the addition theorem of probability in case of non-mutually exclusive events.

Addition Theorem of Probability for Non-Mutually Exclusive Events Statement: If A and B are non-mutually exclusive events, probability of occurrence of either A or B or both is equal to probability that event A occurs, plus probability that event B occurs minus probability of occurrence of events that are common to both A and B.

This can be represented as:

\(P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\)

Addition Theorem of Probability for Non-Mutually Exclusive Events Proof: Let us suppose that a random experiment results in a sample space S with N sample points. Then, by definition:

\(P\left(A\cup B\right)=\frac{n\left(A\cup B\right)}{n\left(S\right)}=\frac{n\left(A\cup B\right)}{N}\)

\(P\left(A\cupB\right)=\frac{\left[n\left(A\right)-n\left(A\capB\right)\right]+n\left(A\cap B\right)+\left[n\left(B\right)-n\left(A\cap B\right)\right]}{N}\)

\(P\left(A\cup B\right)=\frac{n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)}{N}\)

\(P\left(A\cupB\right)=\frac{n\left(A\right)}{N}+\frac{n\left(B\right)}{N}-\frac{n\left(A\cap B\right)}{N}\)

\(P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\)

In General,

If A, B, and C are non-mutually exclusive events, then the probability of occurrence of atleast one event is:

\(P\left(A\cup B\cup C\right)=\frac{n\left(A\cup B\cup C\right)}{N}\)

\(P\left(A\cupB\cupC\right)=\frac{1}{N}\left[n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(B\cap C\right)-n\left(A\cap C\right)+n\left(A\cap B\cap C\right)\right]\)

\(=\frac{n\left(A\right)}{N}+\frac{n\left(B\right)}{N}+\frac{n\left(C\right)}{N}-\frac{n\left(A\cap B\right)}{N}-\frac{n\left(B\cap C\right)}{N}-\frac{n\left(A\cap C\right)}{N}+\frac{n\left(A\cap B\cap C\right)}{N}\)

\(=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\capB\right)-P\left(B\capC\right)-P\left(A\cap C\right)+P\left(A\cap B\cap C\right)\)

Adding Probabilities :

In probability, when you add up the chances of all possible outcomes in an experiment, the total is always 1. This means that something from the possible outcomes must definitely happen.

For Mutually Exclusive Events:

If an experiment has several outcomes that can't happen at the same time, the total of their probabilities is 1.
Example: If there are three possible outcomes A, B, and C, then:
P(A) + P(B) + P(C) = 1

Example:

Imagine 8 teams are playing in the Cricket World Cup, and you're only interested in the chance that India wins.

Let’s say:

• Event A: India wins

• Event B: India does not win

Since only one of these can happen, their total probability is:
P(A) + P(B) = 1
So, if you know P(B), you can find P(A) using:
P(A) = 1 − P(B)

For Non-Mutually Exclusive Events 

When two events can happen at the same time, they are non-mutually exclusive. So, when adding their probabilities, we must be careful not to count the overlapping part twice.

Example:

Let’s say you draw one card from a standard deck of 52 cards.

• Let Event A be drawing a King → There are 4 Kings, so P(A) = 4/52
   
• Let Event B be drawing a Heart → There are 13 Hearts, so P(B) = 13/52
   
• There is 1 card that is both a King and a Heart (King of Hearts), so P(A and B) = 1/52

Addition Rules for Probability :

Let’s say there are two events, A and B. The way we add their probabilities depends on whether they can happen at the same time or not.

Rule 1: Mutually Exclusive Events

If events A and B cannot happen together (they’re mutually exclusive), then to find the probability that either A or B happens, you simply add their probabilities:

P(A or B) = P(A) + P(B)

Example: Getting a Head or a Tail in one coin toss.

Rule 2: Non-Mutually Exclusive Events

If events A and B can happen at the same time (they overlap), then we must subtract the part where both A and B happen together to avoid double counting:

P(A or B) = P(A) + P(B) − P(A and B)

Example: Drawing a red card or a face card from a deck — some cards are both red and face cards.

Solved Examples of Addition Theorem of Probability

Example 1: A card is drawn from a pack of 52 cards. Find the probability that the drawn card is either a diamond or an ace of the club.

Solution 1: Let A: Event of drawing a card of diamond.

Let B: Event of drawing an ace of the club.

The probability of drawing a card of diamond P(A) = 13/52.

The probability of drawing an ace of club P(B) = 1/52.

Since the events are mutually exclusive, probability of drawing a card being a diamond or an ace of club is:

\(P\left(A\cupB\right)=\P\left(A\right)+P\left(B\right)\=\\frac{13}{52}+\frac{1}{52}=\frac{14}{52}=\frac{7}{26}\)

Example 2: A card is drawn at random from a pack of 52 cards. Find the probability that the drawn card is either a club or a queen.

Solution 2: Let A: Event of drawing a card of the club.

Let B: Event of drawing a queen card.

The probability of drawing a card of club P(A) = 13/52.

The probability of drawing a queen card P(B) = 4/52.

As one of the queen is a club, the events are not mutually exclusive.

So, the probability of drawing queen of club = \(P(A\cap B)=1/52\)

\(P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\)

\(=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\)

\(=\frac{16}{52}=\frac{4}{13}\)

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