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# Compound Probability

One or more than one outcomes of an experiment is called event. If any event consists of one and more than one event then it is called Compound Event. Independent event is that event in which the probability of occurrence of one will not affect others, where as in dependent event, occurrence of one affect others. If two events cannot exist at the same point of time, then it is called mutually exclusive events.

### Random Experiment

An experiment that has two or more outcomes which vary in an unpredictable manner from trial to trial when conducted under uniform conditions is called Random experiment.

### Event or Simple Event

The outcomes of a random experiment are called events.

### Compound Event

A compound event consists of two or more simple events. Tossing two dice is a compound event.

 Related Calculators Compound Calculator Calculation of Probability Compound Daily Interest Calculator Compound Inequalities Calculator

## Compound Probability Definition

Compound probability is occurring between two independent events. The compound probability is equal to the probability of the first event multiplied by the probability of the second event. A compound event is defined as the joint occurrence of two or more simple events of a random experiment. In this case, the occurrence of the event depends upon two or more simple events.

Compound probability is the product of two simple events.

Formula for finding probability of independent events.

P(A) · P(B) = P(A and B)

Formula for finding probability of dependent events.

P(A) · P(B following A) = P(A and B)

Formula for finding probability of two mutually exclusive events.

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

1. When the outcomes are decomposable, we say they are compound events.
2. Compound events are also known as composite events.

## Probability of Compound Events

A compound event is an event that consists of more than one single event. 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. Compound events can be classified as mutually exclusive or mutually inclusive.

### Mutually exclusive Events

When two events cannot happen at the same time, they are mutually exclusive events.

If A and B are two mutually exclusive events,
=> P(A or B) = P(A) + P(B)

To find the probability of compound events, we use a rule called multiplication rule.

### Multiplication rule of probability for compound events

If A and B are two events, then the probability that both A and B to take place together, denoted by $P(A\cap B)$ is given by $P(A) P(\frac{B}{A})$ or $P(B)P(\frac{A}{B})$.

Here, $P(\frac{B}{A})$ and $P(\frac{A}{B})$ are conditional probabilities.

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

### Conditional Probability

Conditional Probability of a given sample is the probability if it is known to occur and is denoted by $P(\frac{A}{B})$.

### Solved Examples

Question 1: A card is drawn from a deck of 52 cards, and replaced in the deck. The deck is shuffled and a second card is drawn. What is the probability that both cards are diamond.

Solution:

Step 1:
Total numbers of card in a deck = 52

Step 2:
The probability of drawing a diamond = $\frac{13}{52}$ = $\frac{1}{4}$.

Since the first card drawn is replaced before the second draw, the probability of drawing a diamond on the second try is same.

Step 3:
The probability of drawing two diamonds is given by,

=> P(diamond) * P(diamond) = $\frac{1}{4}$ * $\frac{1}{4}$

= $\frac{1}{16}$.

Question 2: A card is drawn at random from a deck. What is the probability that the card is either an ace or a spade.
Solution:

Step 1:
Total numbers of card in a deck = 52

Number of spade in the deck = 13

Number of ace in the deck = 4

Step 2:
The probability of selecting a card from the deck = $\frac{1}{52}$

The probability that the card is an ace = $\frac{4}{52}$ = $\frac{1}{13}$

The probability that the card is a spade = $\frac{13}{52}$ = $\frac{1}{4}$

=> The probability that the card is an ace or a spade = $\frac{4}{52}$ + $\frac{13}{52}$ - $\frac{1}{52}$

= $\frac{16}{52}$

= $\frac{4}{13}$

=>   The probability that the card is an ace or a spade =  $\frac{4}{13}$ .

Question 3: A toy manufactured by a company consists of two parts A and B. In the process of manufacturing of part A 92 out of 100 are likely to be non-defective. Similarly 89 out of 100 of part B is likely to be non-defective. What is the probability that the toy selected at random is non-defective.

Solution:

Let A denote the event that part A is non-defective and B denote the event that part B is non- defective. Since the selected toy has to be non-defective, both the parts must be non-defective.

P(A) = P(part A is non-defective) = $\frac{92}{100}$

P(B) = P(part B is non-defective) = $\frac{89}{100}$

P(selected toy is not defective) = $P(A\cap B)$ = P(A)*P(B)

= $(\frac{92}{100}) * (\frac{89}{100})$

= 0.82

## Independent Events

When there is no connection between two events, they are said to be Independent. Two events are independent if the occurrence of one of the events does not affect the occurrence of other.

When the events are independent, then the probability that both A and B to take place together, denoted by $P(A\cap B)$, is given by the product of the probabilities of both the events. That is, $P(A\cap B) = P(A)P(B)$

Let us consider a few examples when the events are independent.

### Solved Example

Question: A university has to select a teacher from a list of 100 persons. 75 of them are women and 25 are men. 30 of them know English and remaining do not. What is the probability of selecting English knowing man as the teacher?

Solution:

Let A denote the event of being man and B denote the event of knowing English. Here being a man and knowing English has no connection with each other. So the events A and B are independent

P(A) = P(being a man) = $\frac{25}{100}$

P(Knowing English) = $\frac{30}{100}$

P(Selecting English knowing man as a teacher) = $P(A\cap B)$ = P(A)*P(B)

= 0.25 * 0.30

= 0.075

P(Selecting English knowing man as a teacher) = 0.075

## Compound Probability Examples

Given below are some of the examples on compound probability.

### Solved Examples

Question 1: A bag has 12 mangoes, 8 oranges, if 2 fruits are taken at one by one with replacement, what is the probability of getting mango and orange?
Solution:

Let S be the sample space, n(S) = 12 + 8 = 20

A be the event of getting a mango, n(A) = 12

B be the event of getting an orange, n(B) = 8

P(A) =  $\frac{12}{20}$ = $\frac{3}{5}$

P(B) =  $\frac{8}{20}$ = $\frac{2}{5}$

P(A and B) =  $\frac{3}{5}$ *  $\frac{2}{5} = \frac{6}{25}$

P(Mango and Orange) = $\frac{6}{25}$

Question 2: What is the probability of choosing a queen and seven with replacement?
Solution:

Step 1:
Total numbers of card in a deck = 52

Cards of queen = 4

Cards of 7 = 4

Step 2:
P(Queen) = $\frac{4}{52}$

P(Seven) = $\frac{4}{52}$

P(Queen and Seven) = P(Q) · P(S)

$\frac{4}{52}$ *  $\frac{4}{52}$

= $\frac{16}{2704}$

= $\frac{1}{169}$.

Question 3:

The names of 8 boys and 7 girls from a class are put into a hat. What is the probability that the first 2 names chosen will both be boys?

Solution:

Total number of students = 8 + 7

= 15

P (boys) =  $\frac{8}{15}$

P (girls) =  $\frac{7}{15}$

P (first name chosen will be boy) = P(a) = $\frac{8}{15}$

P (second name chosen will be boy) = P(b) =$\frac{7}{14}$

=> P(both are boys) =  $\frac{8}{15}$ * $\frac{7}{14}$

= $\frac{56}{ 210}$
= $\frac{4}{15}$

=> P(both are boys) = $\frac{4}{15}$

## Compound Probability Problems

Given below are some of the practice problems on Compound Probability.

### Practice Problems

Question 1: A bag has 15 mangoes, 10 oranges, if 2 fruits are taken at one by one with replacement, what is the probability of getting mango and orange?
Question 2:

A box contains 7 blue marbles and 12 yellow marbles, if 2 marbles are taken at one by one without replacement, what is the probability of getting blue and yellow marble?

Ans: 1)  $\frac{6}{25}$ ,  2)  $\frac{14}{57}$

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