PROBABILITY

Basic
Understanding with Case of a Coin

Suppose we are provided with a coin

If we toss it there are likely two results

**HEAD**or

**TAIL**

And possibility of coming each is same.

This whole process can be written in the form probability as

Probability of
getting HEAD or TAIL is 1 out of 2.

{

**No of possible outcomes are 2**( HEAD and TAIL)but to get HEAD as an outcome we are provided with only one HEAD same is the case for TAIL.
In form of probability we write it as

**P(H or T)=1/2**

**Note**–The whole concept of probability runs when possibilty of each outcome is same.

In theoratical form

**General Formula**of probability can be written as**P(O)=(Number of favourable outcomes to O)/Total no of possible outcomes**

*Here no of
favourable outcomes means number of items available of the same type as
outcome desired .

**Illustrative examples**

Q. A bag contains of
a white marble, a blue marble and a red marble.All of same size and shape.Kiran
takes out a marble out of bag without looking into it.What is probability that she
takes out a

1.)White marble

2.)Blue marble

3.)Red marble.

Sol. Here total no
of possible outcomes are 3(There are 3 marbles in bag).

As we are provided with

**P(O)=(No of favourable outcomes to O)/Total no of possible outcomes**

1.)As bag contains only one white marble

No of favourable outcomes ( white marble) are 1.

Threrefore
Probability of white marbles

P(W)=1/3

2.)No of blue marbles in bag=1

Therefore

Probability of blue marbles

P(B)=1/3

3.)No of Red marbles in bag=1

Therefore

Probability of red marbles

P(R)=1/3

__Important Derived Result__
If we add probability of all outcomes

**P(W)+P(B)+P(R)= 1/3 +1/3 +1/3= 1**

**It comes out to be 1.**

It means the sum of probability of all outcomes is 1.

**Case of a Dice**

**Description**

**A dice is a cube in nature made up of six faces of equal shape and size in general and probability of getting each face as outcome is same.**

**Numbers printed on faces are 1, 2, 3, 4, 5, 6**

**Hence total no of possible outcomes are 6.**

**Illustrative Examples**

Q. A dice is thrown once. What is the possibility of getting

1.) a number greater
than or equal to 5.

2.)a number less than 5.

Sol.

Total no of possible outcomes are 6.

1.)

No of possible outcomes which are greater than or equal to
5 are 2 i.e.
5 and 6.

Therefore P(O)=2/6= 1/3

2.) No of possible outcomes with outcomes less than 5 are
4 i.e.1, 2, 3, 4

Therefore P(M)= 4/6 = 2/3

__Important Derived Result__
We saw in previous result that

**probability of getting a number 5 or greater than 5 in a dice is 1/3.**
and

**probability of getting a number less than 5 is 2/3 this can also be written as probability of getting a number****NOT****greater than or equal to 5.**
And sum of two

**P(O)+P(not O)= 1/3+ 2/3= 1.**
Or

**P(not O)=1-P(O)**it can also be written as**P(O’)=1-P(O)**

**where P(O’) is said to be the compliment of P(O).**{Taking P(O) as P(O’) }

**Impossibe Event**

**If we are asked to find the probability of a number or event which does not come under its possible outcomes then it is called impossible event.**

Suppose we are asked to find probability of number

**8**in a dice.
Then as we know there is no number in the dice with number 8
therefore probability

P(8)= 0/6= 0

And such probability is called impossible event.

**Case of A pack of Cards**

**Description**

**A complete pack of cards consist of 52 card with four suits of 13 cards each.**

Four suits are

**1.)Heart****2.)Spade**

**3.)Diamond**

**4.)Club.**

And 13 cards are numbered
as

**Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10 , Jack , Queen, King.****Total no of possible outcomes are 52.**

**Illustrative Examples**

Q. One card is drawn
from a deck of well shuffled cards.Find the probability that drawn card is a

1.)A king

2.)Not a king.

Sol. Total no of
possible outcomes 52(No of cards in deck)

No of favourable outcomes or No of Kings=4

Therefore probability of coming a King is

P(K)= 4/52 = 1/13.

2.)Probability of outcome ‘not a king’

No of such outcomes=Total number of possible outcomes
–Number of kings

K’=52-K= 52-4= 48

P(K’)=48/52= 12/13

And it can be seen

P(K)+P(K’)= 1/13+ 12/13= 1

Hence it concludes that probability of a ‘king’ and ‘not a
king’ is 1.

Basis on this these are said to be the compliment to each
other.

Q. Two athletes Subash and Saurav run in a race.If
probability of Subash’s winning is 0.62 Find probability of Saurav’s winning.

Sol.

Given Probability of Subash’s winning

P(Su)=0.62

Now probability of Saurav winning =Probability of Subash’s
not winning

Or P(Sa)=P(Su’)
{P(Su’)= Probability of Subash’s not winning}

As we know P(Su’)=1- P(Su)

P(Su’)=1- 0.62= 0.38

It means P(Sa)=0.38

Q. A bag contains of
48 identical balls of red color balls and 12 identical balls of yellow colors.

A child was asked to randomly take out a ball without
looking into it. Find probability of outcome as a

1.)Red ball

2.)Yellow ball.

Sol. Total no of
possible outcomes =48+12= 60

1.)
Probability of getting a red ball

P(R)= 48/60= 4/5

2.)
Probability of getting a Yellow ball

P(Y)= 12/60= 1/5

P(R)+P(Y)= 4/5+ 1/5= 1

Q. A box contains 4
blue cards 6 green cards and 5 white cards identical in nature.

Find the probability
of

1.)Blue card.

2.)Green card and

3.)White card

Sol. Total number
of possible outcomes= 4+6+5= 15

1.)Number of Blue cards= 6

Probability of Blue card

P(B)= 6/15

2.)Number of Green cards= 4

Probability of Green cards

P(G)=4/15
=

3.)Number of White cards =5

Probability of White cards

P(W)=5/15

Again P(B)+P(G)+P(W)= 6/15 + 4/15+ 5/15=
15/15= 1

**Case of Two Coins tossed simultaneously**

**Concept**

When two identical coins with one face HEAD and another TALE
tossed at the same time outcomes are

**{H,H}, {H,T}, {T,H}, {T,T}**

Here first term in brackets shows outcome of first coin and
second term of second coin.

**Total no of possible outcomes are 4.**

**Illustrative Examples**

Q. Two coins are
tossed simultaneously. Find the probability of getting at least one TAIL.

Sol. Total no of possible outcomes =4 [ {H,H}, {H,T}, {T,H},
{T,T} ]

Outcome with at least one TAIL =3[{H,T}, {T,H}, {T,T} ]

Therefore Probability

P(T)=3/4

Q. Two coins are tossed simultaneously.Find the
probability of getting two heads.

Sol. Total no of possible outcomes =4 [ {H,H}, {H,T}, {T,H},
{T,T} ]

Outcomes with two HEADS=1[ {H,H} ]

Therefore Probability

P(H,H)=1/4

**Case of Two Dice rolled simultaneously.**

As discussed earlier number on dice faces are 1,2,3,4,5,6.

**When two dices are rolled simultaneously total possible outcomes are 36**listed as

First dice

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

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

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

(4,1) (4,2) (4,3) (4,4)
(4,5) (4,6) Second Dice

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

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

First digit belongs to first dice and second digit belongs
to second.

**Illustrative Examples**

Q. Two dice one blue and other grey are thrown at the same
time.

What is the probability that

1.)
Sum of
numbers on dice is 5.

2.)
Sum of
numbers on dice is 13.

Sol. Total number of possible outcomes are 36

1.) From the arrow possible outcomes list it can easily be
seen that arrow drawn on outcomes shows that
sum of numbers is 5. Such possible outcomes are 4.

Therefore Probability is

P(O)=4/36= 1/9

2)
Number
of outcomes with sum of digits on dice as 13 are zero.

(As
it can be observed that sum of digits on dice can never be greater than 12
)

Therefore
Probability

P(S)=0/36= 0

*************************************** END*******************************************