Difference Between Circular And Linear Permutations

Permutations are the different orders in which a group of objects can be arranged.

The single difference between circular and linear permutations is that circular permutations are the different orders in which a group of objects can be arranged in a circle, whereas linear permutations are the different orders in which a group of objects can be arranged in a straight line.

The formula for circular permutations is obviously not the same as that for linear permutations. For a group of 'n' different objects to be arranged in circular order, the formula is

`P = (n - 1)!`

whereas for the same group of 'n' different objects arranged in linear order, the formula is

`P = n!`

In order to understand the above difference in more detail, read on.

The main difference between a straight line and a circle is that a straight line has a fixed starting point and a fixed ending point, whereas a circle has no fixed starting or ending point. Any point on a circle can be chosen as a starting point and the circle can be drawn from that point.

Thus when you arrange, say, 'n' objects in a straight line, they start with one fixed position and end on a fixed position. On the other hand, when these 'n' objects are arranged in a circle, we can not assign any position to be the starting or ending one. This gives rise to a peculiar property of circular permutations:

When these 'n' objects are arranged in a given order in a circle, and then each object is moved to the position to its right (or left), you get a seemingly different order, but from the mathematical point of view, both orders are same.

For example, the following example shows a group of five differently colored balls arranged in two (seemingly) different circular forms.

Circular Permutations

In the first figure above, the topmost ball is yellow colored while in the second figure, the topmost ball is red. Both the permutations look different, but they are same because in each of the two figures, each ball has the same colored ball to its left and right. For example, consider the yellow colored ball: In both the figures 1 and 2, the ball to its left is red and that to its right is blue. Thus, both the orders above are considered the same.

Now we arrange the same five balls in a straight line.

Linear permutations

The above two orders are considered different because the ball at the starting and ending position are different.

Thus, because a circle does not have a definite starting point, but a straight line does, circular permutations can not be calculated in the same way as linear permutations.

Calculation of Circular Permutations

Circular permutations are calculated by simply fixing one position as the starting and ending position (because the starting and ending points on a circle are always the same). We place any one of the 'n' objects in this fixed position. The remaining (n - 1) objects are now arranged in the remaining (n - 1) positions.

Circular permutations: Fixing one position

 Since we have already fixed one position in the circle, the remaining positions can be considered to be in a straight line, starting from the left of the fixed position and ending on the right of the fixed position (or vice versa).

Thus, in order to calculate the different orders in which the remaining (n - 1) objects can be arranged in the remaining (n - 1) positions, we just need to apply the formula for linear permutations.

`P = (n - 1)!`

The above formula gives the permutations of the (n - 1) objects in the circle's (n - 1) positions. But how does it give us the number of permutations of 'n' objects in the circle?

Now comes in use the concept which we discussed with the first diagram above: In circular permutations, moving all objects by a fixed number of places to their right or left does not change the order. Thus, if we move the fixed position above to any position in the circle (and simultaneously move all the other positions in the circle as well), we will end up with the same order! This is illustrated in the figure below.

Circular permutations: Moving the fixed position gives rise to the same order.

Thus, we can say that the total number of different orders in which 'n' objects can be arranged in a circle are given by

`P = (n - 1)!`

Conclusion

Thus we see that circular permutations are quite different from linear permutations, both in the concept and formulas.

Linear Permutations

Introduction 

When objects are arranged in a straight line, the different orders in which they are arranged are called linear permutations.

For example, four boys and three girls are standing in a straight line. The different orders in which they can stand are represented below: 

• BBBBGGG
• BGBGBGB
• GGGBBBB
• ... and many more 

All of these different orders are called linear permutations.

Formulas

Linear permutations are calculated by different formulas, depending on the type of objects which are arranged in a straight line.

When all objects are different (unique from each other), and all objects are arranged in different orders, linear permutations of these objects are given by: 

`P = n!`

(The above result follows directly from the Fundamental Principle of Counting)

The above formula is explained in more detail in this post.

When all objects are different (unique from each other) and only some of the objects are taken at a time and arranged in different orders, the linear permutations of these objects are given by:

`P_r^n = (n!)/(n-r)!`

(the above formula gives the total number of different orders, that is, different permutations, in which 'n' different objects can be arranged, taken 'r' at a time)

The above formula is explained in more detail in this post.

When there are 'n' objects, out of which 'p' are same and of one kind, 'q' are the same and of a different kind, and 'r' are also same and of a different kind, then the total number of ways in which the 'n' objects can be arranged in different orders are

`P = (n!)/(p!q!r!)`

(in the above formula, the 'p', 'q', and 'r' objects together make the total of 'n' objects. That is, p + q + r = n)

The above formula is explained in more detail in this post.

When there are 'n' objects, alike in all respects, then any order in which they are arranged is the same. Thus, the number of permutations of 'n' alike objects is 1.

Theoretical Probability

When probability of an event is calculated without actually performing the experiment, it is called theoretical probability.

For example, a number cube has six faces, numbered 1 to 6. Out of these six numbers, three are even (2, 4 and 6). Thus, we can write,

Total number of possible outcomes = 6

Number of favorable outcomes (of even numbers) = 3

Now we write the formula for theoretical probability.

`P(A) = "Number of favorable outcomes"/"Total number of possible outcomes"`

Put the values of the number of favorable outcomes and total number of outcomes,

`P("even number") = 3/6`

Simplify the fraction,

`P("even number") = 1/2`

Thus, the probability of getting an even number on rolling a number cube is `1/2`.

In the above problem, we calculated the probability without actually doing an experiment, but with just the concept of the number of favorable events, the total number of events and the formula for theoretical probability.

Thus we see that theoretical probability is simply calculated in three steps:

• Calculate the total number of outcomes in the sample space
• Calculate the number or favorable outcomes in the sample space
• Put these values in the above formula

The following examples will make the concept of theoretical probability clearer.

Problem 1

What is the probability of getting tails on tossing a coin?

A coin has two sides, heads and tails. On tossing it, we can get either heads, or tails. Thus, the sample space is {heads, tails} and the total number of possible outcomes is 2. Further, there is only one "tails" in the sample space. Thus, the number of favorable outcomes is 1. Now we apply the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of possible outcomes"`

 That is,

`P("tails") = 1/2`

Thus, the probability of getting tails on tossing a coin is `1/2`.

Problem 2

What is the probability of getting the number 3 on rolling a number cube?

As explained in the first example in this post, there are six faces on a number cube numbered 1 to 6. On rolling the cube, you can get any one of the six faces facing up. Thus the total number of possible outcomes is 6. Now only one face of the number cube is numbered 3. Thus the number of favorable outcomes is 1. Thus,

`P(A) = "Number of favorable outcomes"/"Total number of possible outcomes"`

Putting the values in the formula,

`P(3) = 1/6`

Thus, the probability of getting the number 3 on rolling a number cube is `1/6`.

Problem 3

What is the probability of getting a prime number on rolling a number cube?

There are four prime numbers on a number cube: 1, 2, 3 and 5. Total number of numbers on it are 6. Thus, the number of favorable outcomes is 4 while the total number of possible outcomes is 6. Thus,

`P("prime number") = "Number of prime numbers"/"Total number of numbers"`

`P("prime number") = 4/6 = 2/3`

Thus, the probability of getting a prime number on rolling a number cube is `2/3`.

Problem 4

What is the probability of selecting a red card from a standard deck of 52 cards?

There are 26 red and 26 black cards in a deck of 52 playing cards. On picking one card, we can get any one of the 52 cards. Thus, the total number of possible outcomes is 52. Since there are 26 red cards, the picking any one of the 26 red cards will result in a favorable outcome. Thus, the number of favorable outcomes is 26.
Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P("red card") = 26/52 = 1/2`

Thus, the probability of getting a red card when picking one card out of a deck of 52 cards is `1/2`.

Problem 5 

There are four red, three yellow and five blue marbles in a bag. What is the probability that you pick a blue marble from the bag?

Total number of marbles in the bag is 4 red + 3 yellow + 5 blue = 12. Picking a marble can result in any one of the 12 marbles, thus the total number of possible outcomes is 12. Further, picking any one of the five blue marbles will result in a favorable outcome. Thus, the number of favorable outcomes is 5. Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P("blue marble") = 5/12`

Probability - P(A or B)

Definition

P(A or B) simply means "The probability of occurrence of event A or event B".

For example, suppose event A is "getting a 4 on rolling a number cube" and event B is "getting heads on tossing a coin", then P(A or B) represents the probability of either of the following three situations:

• You do get 4 on rolling the number cube and you don't get heads on tossing the coin
• You don't get 4 on rolling the number cube but you do get heads on tossing the coin
• You get both 4 on rolling the number cube and heads on tossing the coin

Note that if you don't get 4 on a rolling the number cube and you don't get heads on tossing the coin, then we say that "A or B" has not occurred.

Thus, if either event A occurs or event B occurs, or both the events A and B occur together, then we say that "A or B" has occurred.

P(A or B) is also represented by P(A U B) and P(A + B).

Understand

All the three above expressions, P(A U B), P(A or B) and P(A + B) mean the same thing, "probability of A or B".

For example, you roll a number cube. Then the expression P(1 or 6) means "probability of getting 1 or 6 on rolling the number cube".

How to Calculate

The probability of P(A or B) is calculated either by manually listing the elements in set `AUB` or by the Addition Theorem. Both of these methods are discussed with a sample problem below.

Formula

The addition theorem of probability can be used to calculate P(A or B). It is as follows:

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

Although you can use the above formula to calculate P(A or B), but, as you will see in the examples below, most questions are solved by simply counting the number of events in the set "A or B".

The following example illustrates both methods of calculating P(A or B).

Problem

Calculate the probability P(Even or greater than 4) on rolling a number cube.

Method 1: Manually list the elements in set AUB

The question is asking us to determine the probability of getting a number which is either even or prime on rolling a number cube.

In this method we will list the elements in set AUB (A union B). The given probability is P(Even or 3). Thus we will use the elements of the set "Even U Prime".

Even numbers on a number cube are {2, 4, 6} and the numbers greater than 4 are {5, 6}. Let A = {2, 4, 6} and B = {5, 6}. Then,

AUB = {2, 4, 5, 6}

There are four elements in set AUB. Thus number of favorable outcomes = 4. (these are the outcomes which favor the probability of either an even number or a prime number on rolling the number cube).

We already know that there are six faces on a number cube and hence total number of possible outcomes is 6. Thus, by the definition of theoretical probability,

P(Even or Prime) = `"Number of favorable outcomes"/"Total number of outcomes"`

P(even or prime) = `4/6 = 2/3`

Thus we found that the probability of getting a number which is either even or a prime number is `2/3`.

Method 2: Addition Theorem

The addition theorem of probability helps you calculate the probability of event A or event B. It is as follows.

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

Thus in the above question, we can write,

`P("Even or Prime") = P("Even") + P("prime") - P("even and prime")`

Thus, we calculate each probability P(even), P(prime) and P(even and prime) individually and then plug them in the formula above.

`P("even") = "Number of even numbers on a number cube"/"Total number of numbers on a number cube" = 3/6 = 1/2`

(There are three even numbers on a number cube: 2, 4 and 6.)

`P("prime") = "Number of prime numbers on a number cube"/"Total number of numbers on a number cube" = 4/6 = 2/3`

(There are four prime numbers on a number cube: 1, 2, 3 and 5.)

`P("even and prime") = "Number of numbers on a number cube which are both even and prime"/"Total number of numbers on a number cube" = 1/6`

(2 is the only number which is prime and even.)

Now we plug in the above probability values into the addition theorem.

`P("even or prime") = 1/2 + 2/3 - 1/6`

`P("even or prime") = (3 + 2 - 1)/6 = 4/6 = 2/3`

Thus we calculated the same probability by using the addition theorem.

Some more solved examples are provided below.

Solved Examples

Problem 1

Calculate the probability of getting either an even number or 3 on rolling a number cube.

To solve this problem, first we will find the number of elements in the set "even number or 3". There are six faces on a number cube, numbered 1 to 6. Out of these, three are even: 2, 4 and 6. Thus, including 3, there are 4 faces. Thus, the number of favorable outcomes is 4.

Further, since any of the six faces may show up on rolling the number cube, therefore the total number of possible outcomes is 6.

Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P(even or 3) = 4/6 = 2/3`

Problem 2

There are ten marbles in a bag: 2 red, 3 green, 4 blue and 1 yellow. You pick one marble out of the bag. Find the probability that the marble will be either blue or yellow.

Total number of marbles in the bag = 10. Thus total number of possible outcomes is 10. Further, there are 4 blue and 1 yellow marbles in the bag. Thus, number of favorable outcomes is 4 + 1 = 5.

Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P(blue or yellow) = 5/10 = 1/2`

Problem 3

What is the probability of getting either heads or tails on tossing a coin?

This question is very easy to answer - the probability of getting either heads or tails on tossing a coin is 1. This is because you can get either heads or tails on tossing a coin and there is no other possible outcome. Thus, the probability that you get either heads or tails is 1.

Problem 4

On rolling two number cubes, find the probability that the sum of the numbers you get on both the cubes is either 10 or 12.

First, we list all the possible combinations of numbers on the two number cubes that give a sum of 10:

• 4 + 6 = 10
• 5 + 5 = 10
• 6 + 4 = 10

Now, we list all the possible combinations of numbers which give a sum of 12

• 6 + 6 = 12

Now, we count the number of combinations which give either sum of 10 or a sum of 12. Three combinations give a sum of 10 and one gives a sum of 12. Thus a total of 4 combinations give a sum of 10 or 12. Therefore the number of favorable outcomes is 4.

Now, we calculate the total number of possible outcomes. When rolling a cube, we can get either one of the six faces on it. Thus, there are six possible outcomes on rolling one number cube. When we roll two number cubes, the number of possible outcomes on the second cube for each outcome of the first one are 6. Thus, in total there are (by applying the fundamental principle of counting) `6 times 6 = 36` possible outcomes on rolling two number cubes.

Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P(blue or yellow) = 4/36 = 1/9`

Problem 5

There are three red, four blue and three white marbles in a bag. One marble is selected from the bag. What is the probability that it will be either red or blue?

Total number of marbles in the bag are 3 + 4 + 3 = 10. Thus the total number of possible outcomes is 10. Number of marbles that are either red or blue are 3 + 4 = 7. Thus, the number of favorable outcomes is 7.

Applying the formula for theoretical probability,

`P(A) = "Number of favorable outcomes"/"Total number of outcomes"`

`P(blue or yellow) = 7/10`

Thanks for reading till here (if you did). Hope that it helped you understand P(A or B). Please like, comment and share.

Complementary Events in Probability

Definition

In probability, two events which constitute the total sample space of the experiment are called complementary events. In other words, the combined list of favorable outcomes for both the events equals the sample space of the experiment. For example, on tossing a coin, the sample space is {heads, tails}. The two events P(heads) and P(tails) constitute the sample space of the experiment. Thus they are complementary events.

Notation

If A is an event, then its complement is denoted by A' or "not A". In the above example, the complement of "heads" is "not heads".

Sum of probabilities of complementary events is 1

Since the complementary events together constitute the sample space, thus we can conclude that if one event does not occur, its complement will definitely occur. In other words, either one of the events will definitely occur. Thus the probability of occurring of either one of the events is 1.

Hence, the sum of probabilities of complementary events is 1.

`P(A) + P(A') = 1`

Thus, if we know the probability P(A) of an event A, we can calculate the probability of its complement by using the equation

`P(A') = 1 - P(A)`

For example, we know that the probability of getting 1 on a number cube is `1/6`. Then the probability of getting any of the other five numbers (2 to 6) on a number cube is a the compliment of the above event since it is the same as the probability of not getting 1. Thus

`P(2 " to " 6) =  1 - P(1) = 1 - 1/6 = 5/6`.

Examples

On tossing a coin, the probability of getting heads is

`P("heads") = 1/2`

and the probability of getting tails is

`P("tails") = 1/2`

The sum of their probabilities is

`P("heads") + P("tails") = 1`

Thus the two events "getting heads" and "getting tails" are complementary events.

Similarly, on rolling a number cube, the probability of getting 1 is

`P(1) = 1/6`

The probability of not getting 1 is

`P(1') = 1 - 1/6 = 5/6`

Thus, getting 1 and not getting1 on rolling a number cube are two complementary events.

Complementary events are always Mutually Exclusive

As we know from this post, mutually exclusive events are those events which can not occur together. Complementary events are always mutually exclusive, that is, they can not occur together. For example, on tossing a coin, the result can be either heads or tails, but not both.

Mutually Exclusive Events

Definition

Mutually exclusive events are those events which can not occur together. Mutually non-exclusive events are those events which can occur together because they have one or more common outcomes.

Understand By Example

Let us understand mutually exclusive events by an example.

On tossing a number cube, you can get any one of six faces up. Getting 1 on the cube is an event and getting another number, say 4, is another event. We know that a number cube can't show both the faces 1 and 4 at the same time. Thus, only one of the two events 1 or 4 may occur. In other words, if the outcome is 1, you can't get 4 and if the outcome is 4, you can't get 1.

Such events of an experiment, either of which may occur individually but both may not occur together, are called mutually exclusive events, because the occurrence of one event excludes the occurrence of the other.

In the above example, suppose one event is "getting an even number". There are three even numbers on a number cube, 2, 4 and 6. Another event is "getting prime number". There are four prime numbers on a number cube, 1, 2, 3 and 5 (1 is not a prime number but it is not even, so we are assuming it to be prime in this example). Now suppose you toss the coin and get an outcome of 2. 2 is a prime number and an even number as well. In this case, both the events (getting an even number and getting a prime number) have occurred together. Such events which may occur together because they have one or more matching outcomes (such as 2 in this case), are called mutually non-exclusive events.

Terms in Probability

Let us take an example to learn the few frequently used terms in probability.

Suppose you toss a coin and you need to find the probability of getting heads. Then, the act of tossing the coin to get heads or tails is called an experiment, getting heads or tails from the experiment is called the outcome, the group of possible outcomes or events (heads or tails) from the experiment is called the sample space, and the outcome/event whose probability we are calculating (that is, getting heads), is called the favorable outcome.

Now suppose you toss a coin ten times. Each toss can have two outcomes. Thus there will be ten outcomes in total (although the sample is much greater, having a size of `2^10`). A collection of one or more of these outcomes is called an event. Suppose you are calculating the probability of getting exactly four heads when tossing a coin ten times, then the collection of four outcomes of 'heads' (that is, favorable outcomes) can be called a favorable event. It is used as "The event of getting four exactly heads on tossing a coin ten times." When an event consists of a single outcome, as in tossing a coin only once, then it is called a simple event.

Use of the terms highlighted above is frequent in questions on probability. Thus you need to be familiar with them to understand the questions.

Types of Probability

Probability can be calculated in two ways. Based on this there are two types of probability.

Theoretical Probability

When probability is calculated without actually performing the experiment, the probability is called Theoretical Probability. In this type of probability, you calculate the total number of possible outcomes of an experiment (called the sample space) and the number of favorable outcomes (the kind of outcomes of which you are calculating the probability). Their ratio is called the theoretical probability.

For example, suppose you want to know the probability of getting an even number on rolling a number cube. We know that there are a total of six faces on a number cube and three even numbers on it. Thus the total number of possible outcomes is 6 while the number of favorable outcomes is 3. Thus the theoretical probability is,

P(even number) = `3/6 = 1/2`

Experimental Probability

When the probability is calculated based on the results of an experiment, the probability is called experimental probability. Experimental probability can vary from experiment to experiment, since all experiments may not have the same results.

In the same example above, in order to calculate the probability of getting an even number on rolling a number cube, you will actually perform the experiment for a fixed number of times. Suppose you roll the number cube ten times and get an even number four times. Thus, the experimental probability is,

P(even number) = `4/10 = 2/5`

In general, experimental probability is quite close (or sometimes equal) to the theoretical probability unless the experiment is biased. For example, in the above examples, the theoretical probability is 0.5 while the experimental probability is 0.4 which is a number near to 0.5.

A Bag of Marbles - Probability

There are four red, three yellow and five blue marbles in a bag. What is the probability of selecting a blue marble?

There are a total of 4 + 3 + 5 = 12 marbles in the bag. Out of these, five a blue. Thus the number of favorable outcomes is 5 (since selecting any one of the five blue marbles gives the favorable outcome) and the total number of possible outcomes is 12 (since any one of the twelve marbles can be chosen). Applying the formula for theoretical probability,

`P("blue marble") = "Number of favorable outcomes"/"Total number of possible outcomes"`
`P("blue marble") = 5/12`

Thus the probability of selecting a marble that is blue is `5/12`.

From the same bag mentioned above, you pick one marble, replace it in the bag, and then pick another marble. What is the probability that both marbles will be red?

First find the probability of selecting a red marble from the bag. There are a total of 12 marbles, thus the total number of possible outcomes is 12. Further, there are four red marbles in the bag. Thus there are four favorable outcomes.

`P("red") = "Number of favorable outcomes"/"Total number of possible outcomes" = 4/12 = 1/3`

After selecting the first marble and then replacing it in the bag, you select another marble. Since the first marble was replaced, therefore the total number and number of red marbles is same. Thus the probability of selecting a red marble in the second attempt is the same as that in the first attempt, `1/3`.

Now find the combined probability of both marbles being red.

`P("both red") = P("first red and second red")`

The word 'and' in the above statement implies that since the two events of selecting the marbles are independent events, their combined probabilities is the product of their individual probabilities. Thus,

`P("both red") = 1/3 * 1/3 = 1/9`

Thus the probability of selecting two red marbles by replacing the first marble in the bag is `1/9`.

In the same question as above, the second marble is selected without replacing the first marble in the bag. Then find the probability of getting two red marbles.

This question is different from the above one in that it does not replace the first marble in the bag. Thus, the probability of getting a red marble in the second attempt changes.

The probability of the first marble selected being red is the same as before, `1/3`, since total number of marbles and number of red marbles is the same.

`P("first red") = 1/3`

After selecting the first marble, 11 marbles are left in the bag. Further, if we assume that the first marble selected was red, then 3 red marbles are left in the bag. The probability of getting a red marble from the bag now is

`P("second red") = "Number of red marbles left in the bag"/"Total number of marbles left in the bag" = 3/11`

Now the combined probability of the first marble being red and the second marble being red is the product of their individual probabilities since both are independent events.

`P("first red and second red") = P("first red") * P("second red")`
` = 1/3 * 3/11 = 1/11`

Thus, the probability of getting two red marbles from the bag without replacing the first one is `1/11`.

Probability - Flipping Coins

Flipping A Single Coin

On flipping a coin, you can get either heads or tails. Thus the sample space is {heads, tails} and total number of possible outcomes on flipping a coin is 2.

Suppose you want to find the probability of getting heads on flipping a coin. Since the coin can land on either heads or tails, hence there is 1 favorable outcome - heads. Applying the formula for theoretical probability,

`P("heads") = "Number of favorable outcomes"/"Total number of outcomes" = \frac{1}{2}`

Flipping Two Coins - or One Coin Twice

First, note that flipping two coins together or flipping just one coin twice - both experiments are same.

On flipping two coins, you can get a combination of heads and tails on the two coins. You can get heads on both coins, tails on both coins, heads on the first and tail on the second, and tails on the first and heads on the second. Thus, the sample space is {HH, HT, TH, TT}, where H represents heads and T represents tails.

From the sample space, there are a total of four possible outcomes.

Now suppose you have to find the probability of getting heads on the first coin and tails on the second coin. There is a single outcome in the sample space matching this. Thus there is 1 favorable outcome.

`P("HT") = "Number of favorable outcomes"/"Total number of outcomes" = \frac{1}{2}`

Flipping Three Coins - or One Coin Thrice

As mentioned above, flipping three coins simultaneously or flipping a single coin three times - both are the same experiments.

The possible number of combinations of heads and tails on three coins is naturally greater than those on two coins. The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Thus there are 8 diffferent possible outcomes when three coins are flipped together.

Suppose you have to calculate the probability of getting three tails. There is only one favorable outcome matching this in the sample space. Thus,

`P("TTT") = "Number of favorable outcomes"/"Total number of outcomes" = \frac{1}{8}`

Finding probability in case of flipping a coin (or doing any other experiment) more than 1 times is easier and faster using the binomial theorem.