**What is Probability? Know the Complete Probability Definition Here**

The Probability definition might be confusing, but when you understand it you can easily solve even the toughest problems.

We can define the word ‘Probability’ as – A measure of how likely an event is to occur’.

Probability is the number of ways of achieving success to the total number of possible outcomes.

Conclusively we can say that ‘*Probability is the measure of Uncertainty’*.

To clear the concept of solving Probability Questions, there are also some basic terminologies which need to be considered.

Below you can check all the relative information to dive deep in Probability concepts.

**Basic Terminologies Used in Probability**

Understanding the basic terminologies of Probability is important to start attempting the questions.

Below you we have described importance terms that are considered while attempting a Probability Question.

**Experiment** –

It’s an event or we can say it’s an operation which provides some well-defined outcomes.

**Random Experiment** –

It’s a type of experiment, in which all the possible outcomes are known, but we can not predict the exact output in advance.

The most common examples of Random experiment are –

Rolling an unbiased dice

Drawing a card from well-shuffled cards

Coin flipping

Picking up a colored ball from a bag containing the differently colored ball.

**Event** –

Outcomes that we get after performing an experiment are known as events. There are different types of events which are elaborated below.

**Compound Event** – Combination of two or more elementary events is called a compound event.

**Elementary Event** – All the possible outcomes after repeating a random experiment under exact conditions is known as Elementary Event.

**Mutually Exclusive Events** – Flipping a coin means either head will come, or tail will come. Both these cases are counted in Mutually Exclusive Events.

**Favorable Events** – Total desired outcomes of an elementary event are counted in Favourable Event.

**Independent Events** – If the outcome of one event is not affecting the outcome of other events then both the events are called independent events.

**Important Probability Formulas List**

Here we have listed the important Probability formulas that can help you to solve the problems easily.

Scroll down to check the list.

**Most Probably Cases Discussed in Probability – Dice, Coins, & Cards!**

Enhance your knowledge in Probability here –

**1. **__Probability Problems on Dice__

__Probability Problems on Dice__

This type of Probability Questions is asked on the basis of rolling dice with six sided dots – 1,2,3,4,5 and 6.

When one dice is rolled, the number of possibilities is 6.

When the two dice are rolled together, the number of possibilities is 6*6 = 36.

To check the outcomes of two dice, below we have shared the sample space of Probability.

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

1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |

6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |

Here in the table, the outcomes (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) are known as doublets and the pair (1,2) and (2,1) are different outcomes.

**2. **__Probability Problems on Coins__

__Probability Problems on Coins__

When we throw a coin in the air there are two conditions that can occur. Whether it lands with head or with a tail.

Most of the time questions are asked on Coins problem.

Probability questions on coins can be asked in three different types –

**When only One Coin Flipped –**

In this case, total events will be 2 and Probability P(E) is = 1/2.

**When Two Coins are Flipped –**

The sample space will be – {(H,T), (T,H), (H,H), (T,T)}

If the question is asked “What is the probability that both the coins shows head?”, The possibility of two heads will be decided as P(H,H) = n(E)/n(S)

Putting the values in formula, we can say that the probability when both the coins show head is 1/4.

**When Three Coins are Flipped –**

This time the sample space will be changed as – {(H,H,H), ( T,T,T), (H,H,T), ( H,T,H), (T,H,H), (H,T,T), (T,H,T) ,(T ,T, H)}

Here, we see the probability that both the coins show tails will be P(E) is = 3/8.

**3.**__Probability Problems on Cards__

__Probability Problems on Cards__

Questions are also asked on the basis of a well-shuffled deck of 52 Playing Cards Probability.

Here you can clear all the concepts to attempt the Playing Cards Probability Questions.

The deck of 52 playing cards is divided into 4 suits of 13 cards each i.e. –

Spades (Black Cards)

Hearts (Red Cards)

Diamonds (Red Cards)

Clubs (Black Cards)

Each suit of cards contains an ace, king, queen, jack, 10,9,8,7,6,5,4,3, and 2.

There are total 12 face cards (King, Queen and Jack) in a deck of playing cards.

Let’s clear it with a simple example –

**Question** – Find the probability of a jack drawn from a deck of 52 cards?

**Solution** –

The formula is same to get the Probability i.e.

Number of Favorable Outcomes/Total Number of Possible Outcome

Here we can see that the number of favorable outcomes of ‘a jack’ is 4 out of 52 cards and the total number of possible outcomes is 52.

So, the Probability of drawing a jack is = 4/52.

## Formulas and Quick Tricks for Probability

**Def. of Probability:**Probability is the measure of uncertainty of any event (any phenomenon happened or bound to happen)**Experiment:**Any phenomenon like rolling a dice, tossing a coin, drawing a card from a well-shuffled deck, etc.**Outcome:**The Result of any event; like number appearing on a dice, side of a coin, drawn out card, etc.**Sample Space:**The set of all possible outcomes.**Event:**Any combination of possible outcomes or the subset of sample space; like getting an even number on rolled dice, getting a head/tail on a flipped coin, drawing out a king/queen/ace of any suit.**Probability = (Number of a Favourable outcome) / (Total number of outcomes)**

or,**P = n (E) / n (S)**

where,

P(A) is the probability of an event “A”

n(E) is the number of favourable outcomes

n(S) is the total number of events in the sample space**Odds in Favour of the Event:**Odds in the favor of any event is the ratio of the number of ways that an outcome can occur to the number of ways it cannot occur.**Odds Against the Event:**Odds against any event is the ratio of the number of ways that an outcome cannot occur to the number of ways it can occur.**Probability Range:**0 ≤ P(A) ≤ 1**Rule of Addition:**P(A∪B) = P(A) + P(B) – P(A∩B)**Rule of Complementary Events:**P(A’) + P(A) = 1**Disjoint Events:**P(A∩B) = 0**Independent Events:**P(A∩B) = P(A) . P(B)**Conditional Probability:**P(A | B) = P(A∩B) / P(B)**Bayes Formula:**P(A | B) = P(B | A) . P(A) / P(B)

## Probability Concept

Probability can be defined as a measure of the likelihood for the occurrence of some event. It always lies between 0 to 1.

## Important Probability Terms & Definitions

The following definitions are fairly critical in obtaining a complete understanding of the subject.

**Sample space:**The set of all possible outcomes of an experiment is called sample space. Examples of Sample Space are: In tossing a coin, S={ H, T} and In rolling dice, we have S={ 1, 2, 3, 4, 5, 6}.

**Event:**A subset of the sample space is called an event. It comprises one or more outcomes of the experiment.

**Mutually exclusive events**: Two events are said to be mutually exclusive if the occurrence of one event eliminates the possibility of the occurrence of another. For example, when a coin is tossed once, the occurrence of heads and tails are mutually exclusive as both of them cannot happen simultaneously.

**Collectively exhaustive events:**Events that together cover all possible outcomes are called collectively exhaustive events. For example, when a die is thrown, getting an odd number and getting an even number are two events which, put together, will account for all possible outcomes. These events are called collectively exhaustive.

**Independent events**: Two events are said to be independent if the occurrence of one event does not affect the occurrence of another. For example, when a coin is tossed and a die is thrown simultaneously, the event of getting a head and the number ‘6’ is an independent event.

*Note: If the events are mutually exclusive and collectively exhaustive, then the sum of probabilities of the events will be equal to 1.*

## Probability Formulas and Tricks

Here are some of the important Probability Formulas that you need to understand to solve the problems in no time.

- When we throw a coin, then either a Head(H) or a Tail (T) appears.
- Dice is a solid cube having six faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.
- A pack of cards has 52 cards. It has 13 cards of each suit, namely Spades, Clubs, Hearts and Diamonds. Cards of Spades and clubs are black cards. Cards of Diamonds and hearts are red cards. There are four honours for each suit. These are Aces, Kings, Queens and Jacks. These are called face cards.

Probability of Occurrence of an event: Let S be the sample space and E be an event.

Then, E <= S.

P(E) = n(E) / n(S)

P(S) = 1

0<= P(E) <= 1

P(Ø) = 0

For any events A and B , we have P(A U B) = P(A) + P(B) – P(A ∩ B)

## Theorems of Probability | Probability formulas and tricks

### 1) Addition Theorem

- Addition theorem on probability states that for a given set of events A and B with probabilities P(A) and P(B),
- P (A∪B) = P(A) + P(B) – P (A∩B)
- If A and B are mutually exhaustive events, then P(A∩B) = P(A) x P(B)
- If A and B are mutually exclusive events, then P(A∩B) = 0
- In other words for mutually exclusive events, P(A∪B) = P(A) + P(B)

**Example 1)** A stationery box contains 3 red paper clips, 4 green paper clips, and 5 blue paper clips. One paper clip is taken from the box and then replaced. Another paper clip is taken from the drawer. What is the probability that the first paper clip is red and the second one is blue?

**Solution- **P(red then blue) = P(red) * P(blue)

= (3/12) * (5/12) = 15/144 = 5/48

### 2) Multiplication Theorem

If A and B are independent events, then P(A∩B) = P(A) x P(B) i.e. the probability of simultaneous occurrence of independent events is equal to the product of the individual probabilities for the occurrence of the events.

**Example 2)** A single card is randomly selected from a deck of 52 playing cards. What is the probability of choosing a queen or a heart?

(a) 4/13

(b) 5/12

(c) 12/13

(d) 6/17

**Solution-** (a) 4/13

P(Queen / Heart) = P(Queen) + P(Heart) – P(Queen of Hearts)

= 4/52 + 13/52 – 1/52 = 16/52 = 4/13

### 3) Conditional Probability

The Probability Formulas of the occurrence of an event A, given that event B, has already occurred is given by P (A / B) = P(A n B) / P(B)

**Example 3)** The probability that it is Tuesday and that a Raman is absent is 0.03. SInce there are 5 college days in a week, the probability that it is a Tuesday is 0.2. What is the probability that Raman is absent given that today is a Tuesday?

**Solution**– P(Absent / Tuesday) = P(Tuesday and absent) / P(Tuesday) = 0.03/0.2 = 0.15

**Example 4)** Two dies are thrown simultaneously and the sum of the numbers obtained is found to be 8. What is the probability that the number 5 has appeared at least once?

**Solution**– For the given problem, the sample space S would be all the possible combinations of numbers on the 2 dies. Therefore S consists of 6 × 6 i.e. 36 events.

Event A indicates the combination in which 5 has appeared at least once.

Event B indicates the combination of the numbers which sum up to 8.

A = {(5, 1), (5, 2), (5, 3)(5, 4)(5, 5)(5, 6)(1, 5)(2, 5)(3,5)(4, 5)(6, 5)}

B = {(2, 6)(3, 5)(4,4)(5, 3)(6, 2)}

P(A) = 11/36

P(B) = 5/36

A ∩ B = 2

P(A ∩ B) = 2/36

### 4) Expected value

If a monetary value is associated with each outcome of an experiment, then the expected value from the experiment, if the experiment is performed multiple times, can be computed as the weighted average of the monetary values for all outcomes of the experiment, with the weights being the individual probabilities of the outcomes.

Probability Formulas can be generalised as follows

**Example 5)** Let us say that in a gambling house, betting happens on the throw of a die. We know that the probability of getting any particular number between 1 and 6 (say 5) in this event is 1/6. In the event of a 6, the casino house pays out Rs. 120 to every participant and in the case of any other number, it charges Rs. 30 to every participant. Now, the expected value of the game, if the game is played over a long period of time, can be calculated as

E(G) = (1/6) x 120 + (5/6) x -30 = -5

Hence, if the game is played a long time, the casino will benefit by Rs. 5 for each experiment.