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Addition Rule of Probability

The probability theory is an important part of mathematics. It has many applications. In general, the probability theory deals with the problems related to finding chances of occurring some event.i.e. determining the likelihood of events. Several rules have been associated with the probability theory. Addition rule is one of them. This rule is applicable in problems involving two events. It is applied when we discuss doing one task and finding a probability of two things occurring in that task.

Suppose we have two events named as M and N. Addition rule generally used in the condition when the question is to calculate the probability of A or B, by which one refers that probability of event M occurs or N occurs or both M, N occur together.

In this article, we are going to discuss statement and formula of addition rule and example based on it.

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Definition

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Probability refers to the chance of happening of a particular event. Say for example if we toss a coin, the chance of getting heads is 50%. Thus we say that the probability of heads is 50% or 0.5. In general, the probability of an event is given by the formula:

Probability = $\frac{\text{number of favourable events}}{\text{total number of possible events}}$


The sum of probabilities of all the n events has to be 1. 

If the probability of a specific event is 1, then that event is said to be a certain event. That event is sure to occur. For example, the event ‘sun rises in the east’. We know that this event is sure to occur. The sun is going to rise in the east.

If the probability of a specific event is 0, then that event would never occur. For example, ‘mercury becomes a satellite of earth’. 


If there are n total possible events, and the events are mutually exclusive then the probability of each event would 
be $\frac{1}{n}$

Mutually Exclusive Events

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Mutually Exclusive Events

Two events are said to be mutually exclusive if they can never occur simultaneously. For example, consider the two events while tossing a coin: getting heads and getting tails. These two can never both occur at the same time. Thus they are mutually exclusive events. As stated earlier, if an experiment has n possible outcomes all of which are mutually exclusive, then the probability of each outcome would be $\frac{1}{n}$ and the sum of probabilities of the outcomes would be $1$.

Non-Mutually Exclusive Events

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Non-Mutually Exclusive Events

If two events are such that they can occur simultaneously, then they are said to be non-mutually exclusive events.
For example, consider a throw of a fair 6 sided die. If two events are described as
Event A = An even number comes up and
Event B = A prime number comes up.

Now, we know that the possible even numbers are 2, 4 and 6.
We also know that the possible prime numbers are 2, 3 and 5.
Note that the number 2 occurs in both the events.
Thus it is possible that the number thrown up is both, an even as well as a prime number.
Therefore these two events are non-mutually exclusive events.

The Addition Rule

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Addition rule is applicable when we want to find the compound probability of two or more events. The addition rule is used when the statement describing the event contains ‘either-or’ format. 

Consider two events, A and B that are possible outcomes for an experiment conducted. The probabilities of A and that of B individually are known to us. Then if we wish to calculate the probability of an event such that either A or B or both occur, then we’d use the addition rule. The same concept applies when we have multiple events: $A_1$,$A_2$,$A_3$ … $A_n$. 

However, the rules for mutually exclusive and non-mutually exclusive events are slightly different. Let us take a look at the formula to understand better.

Formula

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If we have two events A and B that form a part of an experiment, and the probabilities of both these events A and B are denoted by P(A) and P(B), then
 
The combined probability of occurrence of event A or B or both is given by:

P(A $\cap$ B) = P(A) + P(B) - P (A $\cap$ B)

Here, the P(A $\cap$ B) refers to the event that BOTH A and B occur. In simpler words,

A $\cup$ B  is A OR B and 

A $\cap$ B is A and B

Now suppose if we have three non-mutually exclusive events, A, B, and C, then their combined probability can be given 
by:

P(A $\cup$ B $\cup$ C) = P(A) + P(B) + P(C) - P(A $\cap$ B) - P(B $\cap$ C) - P(C $\cap$ A) + P(A $\cap$ B $\cap$ C)

Next, if the events are mutually exclusive, then 

P(A $\cap$ B) = P(B $\cap$ C) = P(C $\cap$ A) = P(A $\cap$ B $\cap$ C)=0

Therefore, for mutually exclusive events the probability of occurrence of A OR B OR C would be:

P(A $\cup$ B $\cup$ C) = P(A) + P (B) + P(C)

Thus if there are $\eta$ mutually exclusive events, then their combined probability would be:

P($A_1\ \cup\ A_2\ \cup\ A_3\  \cup\ …. \cup\ A_n$) = P($A_1$) + P ($A_2$) + P($A_3$) + .... + P($A_n$)

Examples

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Example 1: In a group of 90 players, 40 are wearing blue T-shirts, 30 are wearing red T-shirts and 20 are wearing green T-shirts. Find the probability that a person picked at random is wearing a red or a green T-shirt.

Solution:

Total number of persons = 90

Number of people wearing red = 30

Number of people wearing green = 20

Probability of picking a red = $\frac{30}{90}$ = $\frac{1}{3}$

Probability of picking a green = $\frac{20}{90}$ = $\frac{2}{9}$

Since both the events are mutually exclusive, the combined probability of picking a red or a green would be = 
$\frac{1}{3}$ + $\frac{2}{9}$ = $\frac{3}{9}$ + $\frac{2}{9}$ = $\frac{5}{9}$ $\leftarrow$ Answer!
Example 2: Find the probability of drawing an ace or a club card from a pack of well-shuffled playing cards.

Solution:

Total number of cards in the deck = 52

Number of aces = 4

Let A = event that an ace is drawn, then P(A) = $\frac{4}{52}$ = $\frac{1}{13}$

Number of club cards = 13

Let B = event that a club card is drawn, the P(B) = $\frac{13}{52}$ = $\frac{1}{4}$

Number of club ace = 1

Thus, P(A $\cap$ B) = $\frac{1}{52}$

Therefore, probability of drawing an ace or a club card would be:

P(A $\cup$ B) = P(A) + P(B) - P (A $\cap$ B)

P(A $\cup$ B) = $\frac{1}{13}$ + $\frac{1}{4}$ - $\frac{1}{52}$

P(A $\cup$ B) = $\frac{4\ +\ 13}{52}$ - $\frac{1}{52}$

P(A $\cup$ B) = $\frac{17}{52}$ - $\frac{1}{52}$

P(A $\cup$ B) = $\frac{16}{52}$ = $\frac{4}{13}$  $\leftarrow$ Answer!
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