## Probability - Definitions, Formulas, Problems

By: Admin | Posted on: Apr 22, 2020

## Probability

Today we will cover Probability, knowing the probability definitions, tips, and important tips or shortcuts is very helpful to crack this topic. Probability is an important topic for most of the competitive exams. Probability problems are an important part of the QA section of most competitive exams like SBI, IBPS, PO/Clerk, LIC-AAO, SSC, etc. So it becomes relevant to know the right technique of solving these questions. Let us start with our topic Probability.

## Probability Definition and its purpose?

Probability means chances. Probability is the measure of finding how likely an event P(E), is of happening of an event. It tells the purpose of how likely the event will happen or not. Situations in which each outcome is equally likely then we can find the probability of an event using probability formula. Probability is a chance of happening something or we can use a word known as a prediction.

For example: what is the probability of an event such that whether it will rain or not

If the probability that an event will occur is P(E), then the probability that the event will not occur is 1 – P(E). if the probability that one event will occur is X and the independent probability that another event will occur is Y then the probability that both events will occur is XY.

### Probability Definition:

a) An experiment is a situation, which involves chance or probability that leads to results known as outcomes.

b) An outcome for an event is the result of an experiment.

c) An event is one or more outcomes of a random experiment.

d) Probability is the measure of how likely an event is.

With a good understanding of probability definition, one can easily crack the complex problems related to probability.

### Some Basic Terms and Concepts of Probability

How you can learn probability in math?

(i) An experiment

An action or operation for an event resulting in two or more outcomes is called an experiment.

(a) Tossing a coin is an experiment. The two possible outcomes for this experiment are head or tail.

(b) Drawing a card from a pack of 52 cards is an experiment. There are 52 possible outcomes.

Trail: Performing a random experiment for an event is called a Trail.

(ii) Sample space

The set of all possible outcomes of while performing any experiment is called the sample space, denoted by S. Elements of S is called a sample point.

e.g.. (a) In the experiment of tossing of a coin, the sample space has two points corresponding to head (H) and Tail (T) S (H, T)

(b) When we throw a die, then any one of the numbers 1, 2,3,4,5 and 6 will come up. The sample space S(for dice) = {1, 2, 3, 4, 5, 6)

(iii) Event: Any subset of sample space is an event.

eg. (a) If the experiment is done throwing a die which = has faces numbered 1 to 6, then

X= (1, 2, 3, 4, 5, 6).

Then. A = (1,3,5), B (2, 4, 6), the null set, and S itself are some events with respect to S.

What are the null and certain events?

Null event: The null set is called the impossible event. Getting 8 when a die is thrown is called an impossible event.

The entire sample space is known as a certain event.

(iv) Simple event: An event is called a simple event if it is a singleton subset of the sample space S.

eg. (a) When a coin is tossed, sample space S (H, T)

Let A = (H) = the event of the occurrence of head.

and B = (T) = the event of the occurrence of the tail.

Here A and B are simple events.

(b) When a die is thrown, sample space S = (1, 2, 3, 4, 5, 6)

Let A = {5} = the event of occurrence of 5.

B = {2} = the event of occurrences of 2.

Here A and B are simple events.

(v) Compound events it is the joint occurrence of two or more simple events.

Example: The event of at least one head appears when two fair coins are tossed is a compound event ie. A = (HT, TH, HH)

(vi)Equally likely events: The number of simple events is said to be equally likely if there is no reason for one event to occur in preference to any other event.

Example: While drawing Playing cards from a well-shuffled deck of cards there are 52 outcomes as the cards, which are equally likely.

(vii) Exhaustive events: All the possible outcomes taken together in which an experiment can result are said to be exhaustive

Example:  A card is drawn from a well-shuffled pack of 52 cards.

(i) The following events are exhaustive.

(a) The card is black. (b) The card is red.

(ii) The following events are not exhaustive.

(a) The card is the heart. (b) The card is a diamond.

A and B are exhaustive events of the sample space S, then A U B=S

(vii) Mutually exclusive events: If two events cannot occur simultaneously, then they are mutually exclusive.

If A and B are mutually exclusive, then we will write them as A∩B=∅

e.g. In drawing a card from a well-shuffled pack, the following events A = the card is a spade; B = the card is a heart are mutually exclusive.

• In a single toss of a coin, either the head or the tail will appear and not both.
• In a playing of a cubic dice, either an odd number or an even number will turn up and not both.

The following events are not mutually exclusive

(a) The card is a heart. (b) The card is a king.

The card can be king of heart.

viii)  Complementary events: Let A be an event of a random experiment and S the sample space. All the other outcome which are not in A belong to the subset S–A. The event S – Ais called the complement of E.

ix) Sure Event: Since S is a subset of S, S itself is an event and S is called a sure or certain event.

x) Impossible event: Let F be an event of getting more than two heads in tossing two coins simultaneously. F = {} = Ø. So F is an impossible event. Therefore, A sample is a sure event. An empty set Ø is an impossible event.

Before going through the various examples. Let us see important formulae which will help to solve various questions in the examination.

### Important Probability Formulae and Rules

1. P(E)= Number of favorable outcome/Total number of outcome
2. P(E) + P(Ē) = 1
3. For any Event E, probability lies between 0 to 1
4. It involves multiple events
5. It also involves multiplication rules and additional rules for two events defined on the sample space.

Now the question arises that what are some common cases or situations that we study in probability. So here we have the following cases on which most problems are asked.

## Probability in Real life

Probability is the calculation of the occurrence of the event happening or not. There are many real-life scenarios where probability concepts are applied. Below are some common probability applications in real life.

Cases in Probability

1. Dice Problems
2. Coins
3. Cards
4. Ball Or Boxes

How do you calculate the probabilities of the various events?

### 1. Dice:

The dice used here is the one we used to play ‘Ludo, Snake, and Ladders’.

A dice has numbers 1, 2, 3, 4, 5, and 6 are written over its six faces as shown below.

When a dice is thrown the number that appears on the upper face is an event.

The Problems based on dice are mainly of two types

(I) When Only One dice is thrown once and sample spaces :

In such cases, the number rolling on playing dice is either one of them {1,2,3,4,5,6}.
Here assuming, the concerned event of rolling out ‘1’ is 1 only because 1 is written only one face.
hence, the Probability of rolling number 1 is given by  = 1/6

Similarly, Probability of rolling number 2 (or any number between 3 to 6)

Probability of favorable outcome is 1

Total number of possible outcome is 6

Hence the probability of an event is given by   = 1/6

1. What is the probability of getting an even number on the rolling a dice?

Now, considering the event should have an even number which is 2, 4 and 6 on the dice
So, the probability of Concerned Event( Favorable outcome) = 3
Total number of possible out came= 6
So, the Probability of the event is given by = 3/6 = ½

2. What is the probability of getting an Odd number on rolling a dice?

Now, considering the event should have an even number which is 1, 3 and 5 on the dice
So, the probability of Concerned Event( Favorable outcome) = 3
Total number of possible out came= 6
So, the probability of the event is given by = 3/6 = 1/2

3. What is the probability of getting 6 on rolling a dice?

Now, considering the event should have a 6 on the dice
Hence, the probability of Concerned Event (Favorable outcome) = 1
Total number of possible outcome= 6
So, the Probability of the event is given by = /6 = 1/2

4. What is the probability of getting a combination of ‘2’ and ‘6’ on throwing two dices?

Solutions – Now the concerned event should have number ‘2’ and ‘6’ So concerned events = 2{(2,6) and (5,6)}
Total events = 36
⇒ Probability of the event is given by = 2/36 = 1/18

5. What is the probability of getting a sum of ‘9’ on rolling a dice twice?

Solution – Now, the concerned events should have a sum of 9 i.e. Two dice are required. On which adding the numbers on the dice must be 9.

Number on dice 1 + Number on dice 2 = 9
these cases will fulfill condition: (4, 5); (5, 4) and (6, 3), (3, 6)
Total number of possible outcome = 4
and, we know Total Events ‘36’
⇒ Probability of the event is given by = 4/36 = 1/9

### 2. Coins:

The coin is a currency token, which has two faces, one is head and the other is one tail. So, when tossing a coin in the air and when it lands, it might have either a head or tail faces. Coin questions can be classified into three types of categories as shown below:

I. One Coin once:
When a coin is tossed is only once then there can be two possible outcomes either it will be ahead or a tail. In such cases, the total event is 2

Now let us see some event-based questions on the Coin problems.

Question: Consider an event of tossing a coin. What is the probability of getting ahead?
Solutions Concerned event = 1(One head)
Total Event = 2
⇒ P(E) = 1/2

Question: Consider an event of tossing a coin. What is the probability of getting a Tail?
Solutions Concerned event = 1(One Tail)
Total Event = 2
⇒ P(E) = ½

II. Two Coins or One Coin Twice:
When two coins are tossed together or one coin is tossed two times then the following outcomes can be obtained:

Here, ‘H’ = Head; ‘T’ = Tail.

(H, T) shows that on coin 1 it’s Head while on coin 2 its Tail. Here, we have

Total Event = 4

Question: What is the probability of getting at most one head-on tossing a coin?
So, Concerned Event = 3 {(H, T) (T, H) (T, T)}
Total Events = 4
⇒ P(E) = 3/4

### 3. Cards:

There is a total of 52 cards in a deck in which we have four kinds of the symbol used in playing cards. In each set, we have 13 cards. The Description of different symbols are as below:

i) Spade ⇒Black in color (13 in number)♠
ii) Club(Clove)Black in color (13 in number)♣
iii) Heart ⇒Redin color (13 in number)♥
iv) DiamondsRedin color (13 in number)♦

Each of these 4 variants has 13 cards each as a number from 2, 3, 4,…..10 and Jack, Queen, King and Ace(1). There are
1) 26 red cards and 26 black cards = 52 Cards
2) In red cards Heart & Diamonds each of 2, 3 …. 10 andJack, Queen, King, and Ace.

3) Similarly In black cards Spade & Clove each of 2, 3 …. 10 and, Jack, Queen, King, and Ace.

So, in total there are 13 × 4 = 52 cards.

1. One card is drawn:
In such types of questions, a card is drawn from a well-shuffled deck of cards. Here, the Total Events = 52

Question: What is the probability of getting a King of diamond or Queen of Heart in one draw?

Solutions – Here, ‘King of diamond or Queen of Heart’ means that either the card can be the diamond King or Heart Queen. Clearly, there is only one King of diamond and only one Queen of Heart.
So, Concerned Event = 2
Total Event = 52
⇒P(E) = 2/52 = 1/26

2. What is the probability of getting a RED ace in one draw?

Solution:

P(A)=favorable outcome =Total  of Red Ace = 2

P(B) =total number of outcomes =Total number of cards =52

P(E) =Probability of getting a Red ace =2/52

### 4. Ball/Stones/Marble in (Boxes, Bags)

In such questions, a bag or box contains certain Ball/Stones/Marble and some of them (are) drawn.it may be randomly or it  may behave some condition

I. One drawn:

Question  – In a box, there are 8 Black, 7 blue and 6 pink marble. One marble is picked up randomly. What is the probability that marble is neither black nor green?

Solutions – Total Events = 8 (Black) + 7 (blue) + 6 (Pink) = 21
Since the selected marble has to be neither black nor Pink then it’d be Blue and blue marble are 7.

So, P(E) = 7 ⇒ P(E) = 7/21 = 1/3

II. More than One drawn without replacement:

Question – A bag contains 10 red or 10 blue marbles. The probability of drawing two marbles of the same colors.

Solution – Total Events (as 2 marbles are drawn) = 20C2
The marbles drawn can either be both Red color or both blue color and for ‘OR’ event we add up the two events.
So, P(E) = 10C2 (if Red) + 10C2 (if blue)
⇒ P(E) = (10C2 + 10C2)÷ 20C2 = 10 x 9 + 10 x 9 / 20 x 19 = 9 / 19

Question – A bag contains 10 Red and 10 blue marbles. The probability of drawing two marbles of the same colors.

Solution – Total Events (as 2 marbles are drawn) = 20C2
The balls drawn can be both Red color or both blue color. For ‘And’ event we can multiply both events. So, We have two cases here,

The first case – Probability of getting both marbles as blue –

So, P(E) = 10C0 (if red) x 10C2 (if blue)

⇒P(E) = (10C0 x 10C2) ÷ 20C2 = (1 x 10 x 9) / (20 x 19) = 9 / 38

The second case – Probability of getting both marbles as red –

So, P(E) = 10C2 (if black) x 10C0 (if blue)

⇒P(E) = (10C2 x 10C0)÷ 20C2 = 10 x 9 x 1 / 20 x 19 = 9 / 38

Other Miscellaneous questions:

Q 1. Four people are to be chosen at random from a group of 3 men, 2 women and 4 children. What is the probability of selecting exactly two of them being children?

Solution – Total People in the group = 3 men + 2women + 4children = 9
Since, we have to select 4 people from 9 so Total Event = 9C4
And, since we have to select 2 of children so these 2 person have to be from 4 children so Concerned Event for children= 4C2
And, other 2 people will be from either Men or Women (3+2 = 5), their Concerned event = 5C2
⇒P(E) (for all 4 people) = 4C2 ×5C2
⇒ P(E) = (4C2 ×5C2) ÷ 9C4 = (4 x 3 x 5 x 4) x (2 x3) / 9 x 8 x 7 x 6 = 10 / 21

Q. 2. Ravi and Ashwin give exam. Chance of Ravi’s selection is 1/7 and of Ashwin’s selection is 1/5. Find probability of only one of them is selected.

Solution – P( Ravi selecting) = 1/7
P (Ravi not selecting) = 1 – 1/7 = 6/7
P( Ashwin selecting) = 1/5
P (Ashwin not selecting) = 1 – 1/5 = 4/5
Probability of only one of them is selectedP(E) = P( Ravi selecting)× P (Ashwin not selecting) + P( Ashwinselecting)× P (Ravinot selecting) =  (1/7 x 4/5) + (1/5 x 6/7) = 10/35 = 2/7.

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Author: Ashwani Kumar