## What is Compound Probability?

Compound probability can be defined as the probability of the two events which are independent. It can be defined as the multiplication of the probability of two events that are not dependent.

## Compound Event and its Types

A compound event is one that has more than one possible outcome.

Examples of a compound event are:

- Rolling of a 6-sided die to the get sum as 5.
- Drawing an ace from a pack of 52 cards.

There are two types of compound events; one is a mutually exclusive event and the other is a mutually inclusive event.

### Mutually Exclusive Events

Those events which cannot happen together or in other words which cannot occur simultaneously are called mutually exclusive events.

Suppose the two events A and B are treated to be mutually exclusive, the probability of any one of the events A or B will be the total sum of the probability of A and the probability of B.

Example- Tossing a fair coin to obtain either head or tail.

The formula for the probability of mutually exclusive events A or B is given by:

$P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$### Mutually Inclusive Events

Those events which have a non-empty intersection are called mutually inclusive events.

Suppose that there are two events X and Y that are mutually inclusive, the probability of either X or Y will be the difference of the sum of the probability of X and the probability of Y, with the probability of X and Y.

**Example**- Picking an even number and a number less than 5 is a mutually inclusive event because 2 and 4 are even number and less than 5 as well.

The formula for the probability of the mutually inclusive events A and B is given by:

P(A or B) = P (A) + P (B) âˆ’P(A and B)

## Compound Event Formulas

Let A and B be two events in a sample space S and P (A) â‰ 0, P (B) *â‰ *0, then the probabi1ity of event A or B is given by:

P(A or B) = P (A) + P (B) âˆ’P(A and B)

## Multiplication Theorem Compound Probability

If there are two events A and B which are independent, and there is a need to calculate the probability of their simultaneous occurrence or probability of an event, then it will be equal to the product of their individual probabilities.

### Proof Multiplication Theorem

Let n_{1} and n_{2} be the total count of possible cases in events A and B, respectively, to have m_{1} and m_{2} as their respective total cases. Then,

P (A) = Chances of event occurring A = m_{1} / n_{1}

P (B) = Chance of event B = m_{2} / n_{2}

Since, the total number of possible cases in events A and B are n_{1} and n_{2}, respectively and as we are able to combine each of the n_{1} cases possible with event A with each n_{2} case of event B, so the total count of possible cases occurring simultaneously is given by n_{1}.n_{2}. In the same way, the total number of eligible cases that can occur simultaneously for events A and B is given by m_{1}.m_{2}.

Now,

P (Aâˆ©B) = The Chance of the simultaneous occurrence of events A and B

i.e.,

P (Aâˆ©B) = m_{1}.m_{2} / n_{1}.n_{2} = (m_{1} / n_{1})( m_{2} / n_{2} ) = P (A). P (B)

### Note:

- If A, B, and C are three independent events, then the chance of their simultaneous occurrence P (Aâˆ©Bâˆ©C) is given by:

P (Aâˆ©Bâˆ©C) = P (A). P (B) .P (C)

2. The chance that event A will not occur is 1 âˆ’P (A).

Therefore, the probability that no two cases representing A and B will occur is [1 âˆ’ P (A)]. [1 âˆ’P (B)].

Thus, the chances of at least one event occurring = 1 âˆ’ [1 âˆ’ P (A)]. [1 âˆ’ P (B)]

3. Simultaneous occurrence of two independent events A and B can also be listed as P (AB) or P (A and B).

## Conditional Probability

The conditional probability is defined as the probability of the event happening when it is given that the other event has already happened. It is based on the famous Bayesâ€™ theorem.

The conditional probability is based on the relationship between the two events. Let us look at the following example of conditional probability to have a better understanding of the concept.

Suppose the event X is â€˜it is raining outsideâ€™ and its chance is 30% and the event Y is â€˜you need to go outsideâ€™ and its chance is 50%.

Here the solution can be answered by knowing the chance of going outside when it is raining.

### Conditional Probability in Real Life

The conditional probability can be used in many real-life areas and fields.

- Conditional probability will be diversely used as the calculus, politics, and insurance.
- As per conditional probability the chance of re-election of president will depend upon the success of the advertising of the television ads along with the voting preference of the voters. Also, it depends upon the chance of the candidate making the promises during the rallies and the debates.
- Similarly, the chance of rain on a day might depend on other factors such as the formation of rain clouds and the humidity percentage in the air, etc.

### The Formula for Conditional Probability

In conditional probability**:**

P(X | Y) = P (X âˆ© Y)/P(Y)

Where:

- P(X|Y) is the conditional probability. It signifies the chance of event X happening when it is given that event Y has happened already.
- P(X âˆ© Y) is the probability that both events X and Y occur jointly.
- P(Y) is the probability of event Y.

### Formula for the Bayesâ€™ Theorem

$P\left(A|B\right)=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)}$Bayesâ€™ theorem is mainly used in real life in order to predict the actual chance of having a disease when it is given that the test is coming out to be positive.

It is used as a concept for Bayesian learning as a machine learning concept. The stocks of a market can also be predicted by calculating the probability of them using the Bayesâ€™ theorem.

### Conditional Probability for Independent Events

The independent events are those events which in which the happening of the first event does not affect the happening of the second event. For this reason, the conditional probabilities for the two independent events A and B are as follows:

P (A | B) = P (A)

P (B | A) = P (B)

### Conditional Probability for Mutually-Exclusive Events

In the realm of opportunity, the most special event is one that cannot take place at the same time. In other words, if an event has already taken place, another event cannot happen. Therefore, the conditional opportunities for such events always remain a priority

P(A|B) = 0

P(B|A) = 0

## Formulas

- Let A and B be two events in a sample space S, then,

(a) P(A or B) = P (A) + P (B) âˆ’P(A and B)

(b) (X | Y) = P (X âˆ© Y)/P(Y)

(c) $P\left(X|Y\right)=\frac{P\left(Y|X\right)P\left(X\right)}{P\left(Y\right)}$

- The probability of mutually exclusive events A or B is given by:
$P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$.

- If A and B are independent events, then,
- The conditional probabilities are: P (A | B) = P (A) and P (B | A) = P (B).

- The probability of A and B occurring together is: P (Aâˆ©B) = P (A). P (B)

- The probability that none of A and B will occur is: [1 âˆ’ P (A)]. [1 âˆ’ P (B)].

- The probability that at least one of A and B will occur is 1 âˆ’ [1 âˆ’ P (A)]. [1 âˆ’ P (B)].

- If A, B, and C are three independent events, then the chance of their simultaneous occurrence P (Aâˆ©Bâˆ©C) is given by: P (Aâˆ©Bâˆ©C) = P (A). P (B) .P (C).

## Context and Applications

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for:

- Bachelor of Science in Statistics
- Master of Science in Statistics

## Practice Problemï»¿

- Letâ€™s assume a diagnostic test is organized which holds 99% accuracy for covid-19 testing and 60% of the total people have covid-19. We need to find the probability that they actually have the disease given that a patient tests positive.

### Probability Calculation

It is given that the probability of detecting positive when the person has covid-19 is 99%, i.e., 0.99. This implies that P(positive | covid-19) = 0.99.

The percentage of people who have covid-19 is 60%, i.e., P(covid-19) = 0.6.

The probability of detecting positive = 0.6*0.99+0.4*0.01=0.598.

$$\begin{array}{c}P\left(\text{covid19}|\text{positive}\right)=\frac{P\left(\text{positve}|\text{covid19}\right)P\left(\text{covid19}\right)}{P\left(\text{positive}\right)}\\ =\frac{0.99\xc3\u20140.6}{0.598}\\ =0.9933\end{array}$$The following two questions are based on compound probability. Try and solve them in order to calculate the probability of the following two events.

2. Suppose three unbiased coins are tossed, now find the probability that all three heads will appear.

Solution:

Step 1): Write the sample space for the event of tossing three coins.

The sample space for the event of tossing three coins can be written as $\text{S=(TTT,TTH,THT,HTT,THH,HTH,HHT,HHH)}$.

As can be seen, the total number of outcomes is 8.

Step 2): Calculate the probability.

The formula for the probability of an event $E$ is given by $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$, where $n\left(E\right)$ is the number of favorable outcomes to $E$ and $n\left(S\right)$ is the total number of possible outcomes.

The number of outcome that all three heads will appear in a toss of three coins is just 1.

Substitute 8 for $n\left(S\right)$ and 1 for $n\left(E\right)$ to determine the probability that all three heads will appear.

$P\left(\text{threeheads}\right)=\frac{1}{8}$

3. If a pair of dice is rolled, find the probability of obtaining the sum as 9.

Solution:

Step 1): Write the sample space for the event of rolling a pair of dice.

The sample space for the event of rolling a pair of dice can be written as $\begin{array}{l}\text{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}\\ \text{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),}\\ \text{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),}\\ \text{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),}\\ \text{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),}\\ \text{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}\end{array}$.

As can be seen, the total number of outcomes is 36.

Step 2): Calculate the probability.

The formula for the probability of an event $E$ is given by $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$, where $n\left(E\right)$ is the number of favorable outcomes to $E$ and $n\left(S\right)$ is the total number of possible outcomes.

The favorable outcomes for sum of 9 are (3,6), (4,5), (5,4), (6,3) and hence the number of favorable outcomes is 4.

Substitute 36 for $n\left(S\right)$ and 4 for $n\left(E\right)$ to determine the probability of obtaining the sum as 9.

$P\left(\text{sumof9}\right)=\frac{4}{36}$

$=\frac{1}{9}$

### Want more help with your statistics homework?

*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.

### Search. Solve. Succeed!

Study smarter access to millions of step-by step textbook solutions, our Q&A library, and AI powered Math Solver. Plus, you get 30 questions to ask an expert each month.

### Probability and Random Processes

### Conditional Probability, Decision Trees, and Bayes' Theorem

## Compound Probability Homework Questions from Fellow Students

Browse our recently answered Compound Probability homework questions.

### Search. Solve. Succeed!

Study smarter access to millions of step-by step textbook solutions, our Q&A library, and AI powered Math Solver. Plus, you get 30 questions to ask an expert each month.