A compound event describes the probabilistic outcomes from two or more events. Probability of a compound event is combination of probability of two or more simple events. For example, rolling two dices or rolling a dice and tossing a coin. A compound event is defined as a subset of the sample space that contains more than one simple event, and denoted by capital letters. Compound events can be
classified as "**mutually exclusive**" or "**mutually inclusive**".

If two or more events occur simultaneously, they are said to be compound events.Let us go ahead and learn more about compound events, their properties and examples based on them.

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The compound event involves combining two or more events together and finding probability of a particular event or combination of events. The possible outcomes in a compound event are generally more than one. Compound events are probabilities of two or more things happening at once. Some compound events do not affect each other's outcomes, such as tossing a coin and throwing a die. Probability experiments which involve more than one activity are called compound events. Compound events can be
classified as mutually exclusive or mutually inclusive. Finding the probability of a given compound event is done by determining probabilities of individual events and finding their sum and removing overlapping probabilities if there any.

**Mutually exclusive compound events: **Events that cannot happen at the same time are mutually exclusive. Suppose a dice is rolled and the probability that a 1 or a 6 will come is mutually exclusive.

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

**Mutually inclusive compound events:** Events that can happen at the same time are mutually inclusive events. Suppose a card is chosen from a deck of 52 cards, then the probability that it is a black card and a king is mutually inclusive.

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

Compound events are the combination of multiple simple events, such as rolling two dice. In probability of compound events, the word 'and' mean that all parts of the compound events are true and the word 'or' indicate that one or more of the parts of the compound vent is true. The sample space for compound events can be formed using a tree or a table diagram which make it easier to calculate the probability.

The probability of a compound event consisting of two independent events is found by

P(A and B) = P(A) $\times$ P(B)

If the two events are dependent then the probability is calculated as

P(A and B) = P(A) $\times$ P(B following A)

To find the probability of one or the other of two mutually exclusive events.

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

Below you could see some problems based on compound events:

### Solved Examples

**Question 1: **60% of the people in a town are graduates and
40% are non graduates. The probability that graduates getting a job
is 0.5 and the probability that non graduates getting a job is 0.20.
One person is selected at random. What is the probability that the
selected person gets a job?

** Solution: **

P(selected person gets a job) = P(graduate is selected and he gets job or non graduate is selected and he get a job)

Let A denote the event that graduate is selected

B denote the event that non graduate is selected

$\frac{C}{A}$ denote the event that the graduate selected gets a job.

$\frac{D}{B}$ denote the event that the non-graduate selected gets a job.

P(A) = P(graduate is selected) = 0.6

P(B) = P(non graduate is selected) = 0.4

P ($\frac{C}{A}$) = P(graduate selected gets a job) = 0.5

P($\frac{D}{B}$) = P(non-graduate selected gets a job) = 0.2

So Required probability is P(A).P ($\frac{C}{A}$) + P(B).P($\frac{D}{B}$)

= (0.6) x 0.5 + (0.4) x 0.2 = 0.3 + 0.08 = 0.38

**Answer: P(selected person gets a job) = 0.38**

**Question 2: **A students scores good marks with probability 0.6 if the test is easy and
0.2 if the test is tough. For series A there is a 0.7 probability of
test to be easy and 0.2 probability for the test to be tough. What is
the probability that the student will score good on series A.

** Solution: **

Let A denote the event that the test is easy, $\frac{B}{A}$ denote the event that he scores good on an easy test on series A. Let C denote the event that the test is tough, $\frac{D}{C}$ denote the event that he score good on a tough test on series A.

P(A) = 0.6

P ($\frac{B}{A}$) = 0.7

P (C) = 0.2

P($\frac{D}{C}$) = 0.2

P[He will score good on series A]

= P (A). P ($\frac{B}{A}$) + P(C) .P ($\frac{D}{C}$)

= 0.6 $\times$ 0.7 + 0.2 $\times$ 0.2 = 0.42 + 0.04 = 0.46

**Answer: P[He will score good on series A] = 0.46**

The compound event can be measured using tree diagrams which provide us a more clear picture of the situation. Let us have a look at an example of rolling a die and tossing a coin together. The tree diagram of possible outcomes may be shown as under :

or as the following diagram too :

Here, we are denoting each and every possible outcome in the form of a tree. Eventually, the possible outcomes of first event are demonstrated by first set of "branches" and similarly, second set denotes possible outcomes of second event. The order of the events doesn't really matter as far as the total number of outcomes are same.

In above example, there are total 12 outcomes. From diagram, we can read easily and find the probability of any particular event. For example - the probability of getting a head and an odd number would be $\frac{3}{12}$ = $\frac{1}{4}$

**Problem 1:** A bookshelf has 5 history books on first shelf and five maths book on second shelf. If Alia chooses one book from each shelf then what is the probability that history book is first from right and maths book is first from left?

**Problem 2:** If one card is chosen from a deck of 52 cards and a coin is tossed, find the probability that one king of black color and head comes together.

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

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

Compound events are the combination of multiple simple events, such as rolling two dice. In probability of compound events, the word 'and' mean that all parts of the compound events are true and the word 'or' indicate that one or more of the parts of the compound vent is true. The sample space for compound events can be formed using a tree or a table diagram which make it easier to calculate the probability.

The probability of a compound event consisting of two independent events is found by

P(A and B) = P(A) $\times$ P(B)

If the two events are dependent then the probability is calculated as

P(A and B) = P(A) $\times$ P(B following A)

To find the probability of one or the other of two mutually exclusive events.

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

Below you could see some problems based on compound events:

P(selected person gets a job) = P(graduate is selected and he gets job or non graduate is selected and he get a job)

Let A denote the event that graduate is selected

B denote the event that non graduate is selected

$\frac{C}{A}$ denote the event that the graduate selected gets a job.

$\frac{D}{B}$ denote the event that the non-graduate selected gets a job.

P(A) = P(graduate is selected) = 0.6

P(B) = P(non graduate is selected) = 0.4

P ($\frac{C}{A}$) = P(graduate selected gets a job) = 0.5

P($\frac{D}{B}$) = P(non-graduate selected gets a job) = 0.2

So Required probability is P(A).P ($\frac{C}{A}$) + P(B).P($\frac{D}{B}$)

= (0.6) x 0.5 + (0.4) x 0.2 = 0.3 + 0.08 = 0.38

Let A denote the event that the test is easy, $\frac{B}{A}$ denote the event that he scores good on an easy test on series A. Let C denote the event that the test is tough, $\frac{D}{C}$ denote the event that he score good on a tough test on series A.

P(A) = 0.6

P ($\frac{B}{A}$) = 0.7

P (C) = 0.2

P($\frac{D}{C}$) = 0.2

P[He will score good on series A]

= P (A). P ($\frac{B}{A}$) + P(C) .P ($\frac{D}{C}$)

= 0.6 $\times$ 0.7 + 0.2 $\times$ 0.2 = 0.42 + 0.04 = 0.46

or as the following diagram too :

Here, we are denoting each and every possible outcome in the form of a tree. Eventually, the possible outcomes of first event are demonstrated by first set of "branches" and similarly, second set denotes possible outcomes of second event. The order of the events doesn't really matter as far as the total number of outcomes are same.

In above example, there are total 12 outcomes. From diagram, we can read easily and find the probability of any particular event. For example - the probability of getting a head and an odd number would be $\frac{3}{12}$ = $\frac{1}{4}$

There is another way for pictorial representation of compound event. This is known as area model which is a chart denoting all the possible outcomes. If we take same example considered in above section, the area model for this would look like the following table -

Here, the columns denote possible outcomes on rolling a die, while rows represent outcomes on tossing a coin.

An area model is drawn by constructing a table which lists the possible outcomes of one event on the top and that of another event on the side. Now, the corresponding cells should be filled with outcomes for that particular event. We may even highlight the cells showing our probability by shading them.

1 | 2 | 3 | 4 | 5 | 6 | |

Tail | T1 | T2 | T3 | T4 | T5 | T6 |

Head | H1 | H2 | H3 | H4 | H5 | H6 |

An area model is drawn by constructing a table which lists the possible outcomes of one event on the top and that of another event on the side. Now, the corresponding cells should be filled with outcomes for that particular event. We may even highlight the cells showing our probability by shading them.

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