Events in a probability experiment can be classified as Simple and Compound events. When an event cannot be broken into smaller events it is known as a Simple event. Simple events have only one outcome. This against the outcomes for a compound event is not restricted to a single outcome. Compound events are also known as multiple events as they can be viewed as the simultaneous occurrence of more than one event. In real world experiences, we require to deal with Compound events more often than with Simple Events. For this purpose specific rules/formulas are derived for computing the probability of compound events.
Let us see, how compound events are defined and how their probabilities are computed.

## Compound Events Definition

An event that consists of two or more simple events is called a compound event.For example, when a die is thrown, the event of number 1 turning up is a simple event.
The event of getting an odd number consists of three simple events of getting 1, getting 3 and getting 5.
Compound events can be defined either as union or as intersection of two events.
Formulas to compute the probability of compound events are derived based on the whether the events
are mutually exclusive, mutually inclusive or independent.

Rule 1:
When two events A and B are mutually exclusive, the probability that A or B will occur is
P(A or B) = P(A) + P(B)

Rule 2:
If A and B are not mutually exclusive (meaning they are mutually inclusive) then
P(A or B) = P(A) + P(B) - P(A and B).

### Multiplication Rule for probability

When A and B are independent events, the probability of both happening is
P(A and B) = P(A) . P(B).

## Compound Events Examples

### Solved Examples

Question 1: Two dice are rolled. Find the probability of getting
a.  A sum of 5, 7 or 9
b. Doubles or a sum of 6 or 8
c. A sum greater than 9 or less than 4.
Solution:

a. The sample space of rolling two dice consists of 36 outcomes.
n(s) = 36.
If E is the event of getting a sum of 5, 7 or 9, then
E = {(1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (3, 6), (6, 3), (5, 4), (4, 5)}
n(E) = 14
Probability ( a sum of 5, 7 or 9) =  $\frac{n(E)}{n(S)}$ = $\frac{14}{36}$ = $\frac{7}{18}$.

b.  If A is the event of getting a double, or a sum of 6 or 8, then
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 5), (5, 1), (2, 4), (4, 2), (2, 6), (6, 2), (3, 5), (5, 3)}
n(A) = 14
P(Doubles or a sum of 6 or 8) = $\frac{n(A)}{n(S)}$ = $\frac{14}{36}$ = $\frac{7}{18}$

c.  If B is the event of a sum greater than 9 or less than 4, then
B = {(1, 1), (1, 2), (2, 1), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
n(B) = 9
Probability (sum greater than 9 or less than 4) = $\frac{n(B)}{n(S)}$ = $\frac{9}{36}$ = $\frac{1}{4}$

Question 2: In a clearance sale of Slacks and T-shirts, the Sales Manager estimates the probabilities of a customer making a purchase worth more and less than 200 dollars are 40% and 30% respectively. Find the probability of a customer not buying anything.
Solution:

The events of a customer buying for more than 200 dollars, less than 200 dollars and nothing are mutually exclusive events.
P(More than 200 dollars) + P(less than 200 dollars) + P(no Purchase) = 1
0.40 + 0.30 + P(no Purchase) = 1
P(no Purchase) = 0.30 = 30%.

Question 3: A survey on School Children's attitude towards studies claimed that 28% of the students hate Math. If three students are selected at random what is the probability that all three hate math.
Solution:

Here a student like or dislike for Math does not depend on another student's attitude towards the subject.
So, we deal with independent events in assessing the probability of all the three students hating Math.
P(hate Math) = 0.28
By Multiplication rule for probabilities
P(All the three students hate Math) = P(hate Math) . P(hate Math) . P(hate Math)
= 0.28 x 0.28 x 0.28 = 0.021952

### Mutually Exclusive Events

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