## Probability:

Probability is a concept that numerically measures the degree of certainty of the occurrence of the event.

**A basic term used in Probability:**

###

**[1] Random Experiment:**

An experiment whose outcome has to be among a set of events that are completely known but whose exact outcome is unknown is a random experiment.By a trial, we mean performing a random experiment.

**Examples:**

- Tossing a fair coin. (We know that by tossing a coin we get either
*H*or*T*but we do not know which one comes*H or T.*) - Drawing a card from a pack of well-shuffled cards.
- Rolling an unbiased die
- Picking up a ball from a bag of balls of different colours.

### [2] Sample Space:

This is defined in the context of a random experiment and denotes the set representing all the possible outcomes of the random experiment.

**Example:-**

- Sample space when a coin is tossed is [Head, Tail]
- Sample space when a dice is thrown is [1,2,3,4,5.6]

###
**[3] Event:**

The set representing the desired outcome of a random experiment is called the event. Note that the event is a subset of the sample space.

**Examples:**

- In throwing a coin,
*H*is the event of getting head. - Suppose we throw two coins simultaneously and let
*E*be the event of getting at least one head. Then,*E*contains*HT, TH, HH*

**[4] Equally Likely Event:**

If two events have the same probability or chance of occurrence they are called an equally likely event. (In a throw of a dice, the chance of 1 showing on the dice is equal to 2 is equal to 3is equal to 4 is equal to 5 is equal to 6 appearing on the dice)

###
**[5] Sure Event:**

It is evident that in a single toss of the die, we will always get a number less than 7.

So, getting a number less than 7 is a sure event.

P(getting a number less than 7) =6/6=1

Thus, the probability of a sure event is 1.

###
**[6]Impossible Event:**

In a single toss of a die, what is the probability of getting a number 8?

we know that in tossing a coin,8 will never come up.

So, getting 8 in an impossible event.

P(getting 8 in a single throw of a die)= 0/6=0

Thus, the probability of an impossible event is zero.

### [7] Mutually Exclusive Event:

A set of an event is a mutually exclusive event when the occurrence of any one of them means that the other events can't occur. (If head appears on a coin, the tail will not appear and vice versa)

.

### [8] Exhaustive Set of Event:

A set of events that includes all the possibilities of the sample space is said to be an exhaustive set of events. (Eg. In a throw of dice the number is less than three or more or equal to three.)

### [9] Independent Event:

An event is described as such if the occurrence of an event has no effect on the probability of the occurrence of another event. (If the first child of a couple is a boy, there is no effect on the chances of the second child being a boy)

### [10] Conditional Probability:

It is the probability of the occurrence of event

*A*given that event*B*has already occurred. This is denoted by*P(A/B).*(**Eg.**The probability that in two throws of dice we get a total of 7 or more, given in the first throw of the dice the number 5 had occurred).###

[11] Complimentary Event:

Let

*E*be an event and (not*E*) be an event that occurs only when E does not occur.
The event (not

*E*) is called the complementary event of*E.*
Clearly,

*P(E)+P(not E)=1*

*∴ P(E) = 1 - P(not E) or*

*P(E)=1 - P(E')*

*P(not E)=*Complementary event denoted by

*E'*

**Note-**

**1**. For some event E, we have

*0≤ P(E)≤ 1.*

**2**

*when*

**.**P(E)=0,*E*is an impossible event.

**3.**

*P(E) = 1*,when

*E*is a sure event .

**4**.

*P(not E) = 1 - P(E)*

### Probability of occurrence of an Event:

The probability of occurrence of an event E, denoted by P(E) is defined as:

P(E)=number of outcomes favourable to ETotal number of possible outcomes

### Some details about these experiments:

- Tossing a coin:- When we throw a coin, either head
*(H)*or a tail*(T)*appears on the upper face. - Throwing a die:- A die is a solid cube, having 6 faces, marked 1,2,3,4,5 and 6 or having 1,2,3,4,5 and 6 dots.
- Drawing a card from a well-shuffled deck of 52 cards.

**(i) It has 13 cards each of four suits, namely**

*spades, clubs, hearts, and diamonds*

*(a) Cards of spades and clubs are black cards.*

(b) Cards of hearts and diamonds are red cards.

**(ii) Kings, queens, and jacks (or knaves) are known as face cards. Thus, there are in all 12 face cards.**

Face cards of diamond |

The underlying factor for real-life estimation of the probability:

### (a) The use of the conjunction AND:

We use AND as the natural conjunction joining two separate parts of the event definition, we replace the AND by the multiplication sign.

Thus, If

*A*AND*B*have to occur, and if the probability of their occurrence is*P(A)*and*P(B)*respectively, then the probability that*A*AND*B*occur is got by connecting*P(A)*AND*P(B)*as,*⟶ P(A) x P(B)*

**Example:**If we have the probability of

*A*hitting a target as 1/3 and that of

*B*hitting the target as 1/2, then the probability that both hit the target if one shot is taken by both of them is by

**Ans: -**A hits the target AND B hit the target

⟶ P(A) x P(B)=1/3 x 1/2 =1/6

### (b) The use of the conjunction OR:

We use OR as the natural conjunction joining two separate parts of the event definition, we replace the OR with the addition sign.

Thus, if

*A*OR*B*has to occur, and if the probability of their occurrence is*P(A)*and*P(B)*respectively, then the probability that A OR B occurs is got by connecting.*P(A)*OR

*P(B)*replacing the OR by addition sign, we get the required probability as

*⟶ P(A) + P(B)*

**Example:-**If we have the probability of A winning a race as 1/3 and that of B winning the race as 1/2, then the probability that either A or B wins a race got by

*⟶P(A)+P(B)=1/3+1/2=5/6*

### (c) Combination of AND and OR:

If two dice are thrown, what is the chance that the sum of the number is not less than 10?

**Event Definition:**The sum of the numbers is not less than 10 if it is either 10 OR 11 OR 12.

(6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR

( 5AND 6) OR (6AND6)

(1/6X1/6)+(1/6X1/6)+(1/6X1/6)+(1/6+1/6)+(1/6X1/6)+(1/6X1/6)

=1/6