The word PROBABILITY is used to indicate an unclear possibility that something might happen. It is also used simultaneously with chance.

### Definition of Probability

If an event ‘E’ can happen in ‘m’ ways & fail in ‘k’ ways out of a total of ‘n’ ways, & each of them is equally likely, then the probability of the happening of ‘E’ is .

P(E) = m/(m+k) = m/n, where n = (m+k)

P(E) = m/(m+k) = m/n, where n = (m+k)

Probability always lies between 0 to 1. If your answer exceeds 1 then your answer is incorrect.

**In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes**

*Experiment*:**: In probability theory, an outcome is a possible result of an experiment. Each possible outcome of a particular experiment is unique**

*Outcome***: The well-defined set of possible outcomes is known as the sample space.**

*Sample Space*### Random Experiment

A random experiment is an experiment or a process for which the outcome cannot be predicted with certainty.

Example: - Drawing 2 cards from a well shuffled pack is a random experiment while getting an Ace & a King are events.

### Mutually Exclusive Events

In probability theory, two events) are mutually exclusive if they cannot both be true or occur at the same time.
Example: The events of getting a head or a tail when a coin is tossed are mutually exclusive.

###

Equally Likely Events

Example: - When a die is thrown, any number from 1 to 6 may turn up. In this trial, the six events are equally likely.

### Independent Events

Two events E1 and E2 are said to be independent, if the occurrence of the event E 2 is not affected by the occurrence or the non-occurrence of the event E 1 .
To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.

*Multiplication Rule 1: When two events, A and B, are independent, the probability of both occurring is:*

*P(A and B) = P(A) · P(B)*

### Compound Events

Example: - When a die is thrown and a coin is flipped the occurring events are called compound events.

## Exercise with Explanation

## Example 1

Find the probability of getting a head in a throw of a coin.

**Solution:-**

When a coin is tossed we either get head or tail upwards.

So, total number of cases= 2 = n,

number of favorable cases to get H = 1 = m

P (H) = No. of favorable cases/ Total no. of outcomes = (m/n)

=1/2

## Example 2

An unbiased die is rolled. Find the probability of a) Getting a multiple of 3 b) getting a prime number

**Solution:-**

When a die is rolled we can get any one of the numbers from 1 to 6.

Total number of cases = n = 6

a) Let event A= getting a multiple of 3

Then A= {3.6}.

Therefore m=2

P (A) = m/n = 2/6 = 1/3

b) Let event B = getting a prime number

Then B= {2, 3, 5}.

so, m = 3

P (A) =m/n=3/6=1/2

## Example 3

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that

a) Card drawn is red

b) Card drawn is Queen

c) Card drawn is black & king

d) Card drawn is red & number card

e) Card drawn is either king or queen

**Solution :**

Before solving this problem, let us recall the game of cards. One deck of cards contains totally 52 cards. Among them we have 13 spades & 13 clubs which are black in color, 13 diamonds & 13 hearts all of which are red. In the 13 spades, 9 are numbered cards, numbered from 2 to 10, one ace card & 3 face cards namely J, Q & K. Similarly for the clubs, diamonds & hearts also.

Totally there are 52 cards & any one can be drawn

So, total number of cases = n = 52

a) There are 13 diamonds & 13 hearts which are red

Number of red cards = m = 26

P (getting red) =26/52=1/2

b) There are 4 queens = 4

P (getting queen) = m/n=4/52

=1/13

c) The king of spade & clubs are black

No. of cards which is king & black=m=2

P (king & black) =m/n =2/56

=1/26

d) The 9 number cards of hearts & 9 number cards of diamonds are red.

No. of cards which are red & number cards = m = 18

P (red & number)=m/n=18/52

=9/26

e) There are 4 queens & kings

No. of favorable case = m = 8

P (queen or king) = m/n = 8/52

= 2/13

## Example 4

A bag contains 6 white beads & 4 red beads. A bead is drawn at random. What is the probability that the bead drawn is white?

**Solution:**Total no. of beads in the bag = 6+4 =10. n= 10

Any one of the 6 white beads can be selected, m=6

Therefore, P (getting white bead) = m/n=6/10

=3/5

## Example 5

A box contains 8 red marbles, 6 green marbles & 10 pink marbles. One marble is drawn at random from box. What is the probability that the marble drawn is either red or green?

**Solution:-**

Total number of marbles= 8+6+10 =24, n=24

There are 6 green & 8 red marbles

Therefore, number of favorable cases=6+8=14

P (red or green) = 14/24=7/12

## Example 6

Two fair coins are tossed simultaneously. Find the probability of

a) Getting two heads

b) at least one head c) exactly one head

**Solution:-**

Sample space S= {TT, TH, HT, HH}

No. of total cases=n=4

a) Getting two heads is only one case i.e. HH

No. of favorable case=m=1

P (getting 2 heads)=m/n= 1/4

b) At least one head means one or more heads. In this case it is one or 2 heads, in 3 cases i.e. TH,HT,HH

Therefore, P (atleast one head) =3/4

c) Exactly one head is two cases TH, HT

P (exactly one head) =2/4=1/2