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Page 2

3

Chapter 2 Mathematics

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30
th

Cross, 10
th

Main, Jayanagar 4
th

Block, Bangalore-11

CHAPTER 2

Probability and Distribution

1. Permutation
It is the arrangement of the given things (objects) in some definite order.
Cases
a) Different things: No. of permutations of n different things taken all at a time = n!

Example

No. of permutation of word “GATE” = 4!

b) Classes of equal things: No. of permutations of n objects of which are alike, are
alike and

are alike =
!

! ! !

c) The number of different permutations of n different things taken r at a time

without repetition = nPr=
!

!

d) The number of different permutations of n different things taken r at a time with
repetition =

Example

In a coded telegram, the letters “a, b, c” are arranged in group of two letters called words. How
many different such words can be coded?

Solution

Given 3 letters are a, b, c. Select 2 with repetitions → 3 = 9 (ab , ba ,ac, ca, bc, cb, aa, bb, cc)

In this problem, number of different such words containing each letter no more than
once(without repetition) is 3P2=6 (ab , ba ,ac, ca, bc, cb)

Example

In a coded telegram, the letters “a, b ,c…………….z ” are arranged in group of 5 letters called
words. How many different such words can be coded?

Solution

Given 26 letters a, b ,c………………..z → select 5 with repetitions

Number of such words: 265 =11881376

No. of different such words containing each letter no more than once is 26P5=
!

!
=7893600

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Chapter 2 Mathematics

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30
th

Cross, 10
th

Main, Jayanagar 4
th

Block, Bangalore-11

18. Continuous Cumulative Distribution function (cdf) or Distribution Function

If ( ) = P(X x) =∫ ( )

then F(x) is defined as the cumulative distribution

function or simply the distribution function of the continuous variable .

CDF has the following properties

i)
( )

= ( ) =f(x) 0

ii) 1 ( ) 0
iii) If ( ) ( )

In other words, CDF is monotone (non-decreasing function)
iv) ( ) = 0
v) ( ) = 1

vi) P(a b) =∫ ( )

= ∫ ( )

- ∫ ( )

= ( ) ( )

Example

The numbers of heads which turn up on tossing three coins is given by the following table.

X= 0 1 2 3

P( )

There are 8 equally likely events = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Check: The sum of all probabilities is equal to one.

Example

The discrete r.v and its probability density function are given below

Xi 0 1 2 3 4 5 6

P (Xi) k 3k 5k 7k 9k 11k 13k

(A) k =? (B) P [ x < 4 ] =? ( C ) P [3 < x 6 ] =?

Solution

(A) ( ) = 1

k + 3k + 5k + 7k +9k + 11k + 13k = 1

49 k =1

k =

(B) P [ x < 4 ] = p [x = 0] +p( x= 1 ) +p(x =2)+ p (x=3)

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Chapter 2 Mathematics

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30
th

Cross, 10
th

Main, Jayanagar 4
th

Block, Bangalore-11

= k + 3k+5k+7k =

(C) P [ 3 < x 6 ] = 9k + 11k + 13k

= 33k =

Example

Consider dice experiment and the random variable X is outcome of the experiment. Sketch the
CDF.

Solution

Note: For a discrete r.v. , the CDF is staircase waveform.

Example

The pdf of a continuous random variable ( ) = 3 0 < x < 1 . Find ‘ a ‘ & ‘ b ‘ such that

X = 1 2 3 4 5 6

P(x) =f(x

)

Fx( 1 ) = P( x 1 ) = 1/6

Fx( 2 ) = P( x 2 ) = 2/6

Fx( 3 ) = P ( x 3 ) =3/6

Fx( 4 ) = P ( x 4 ) =4/6

Fx( 5 ) = P ( x 5 ) =5/6

Fx( 6 ) = P ( x 6 ) =6/6 = 1

y

x
1

3 2 4 5 6

1

5/6

4/6

3/6

Fx(x) =

u(x+1)+

u(x+2)+

(x+3)+

(x+4)+

u(x+5)+

u(x+6)

2/6

1/6

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Chapter 2 Mathematics

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30
th

Cross, 10
th

Main, Jayanagar 4
th

Block, Bangalore-11

Properties of Joint distribution function/ cummulative distribution function:

1. F ( , ) = 0

2. F ( , ) = 1

3. F ( , ) = 0 { F ( , ) = P(X Y y) = 0 x 1 = 0 }

4. F (x, ) = P(X x Y ) = F (x) . 1 = F (x)

5. F ( , y) = F (y)

32. Joint probability density function

Defined as f(x, y) =

F(x, y)

Property: ∫ ∫ f(x, y) dx dy

= 1

Note: X and Y are said to be independent random variable

If fxy(x,y) = fx(x) . fY(y)

Example

For a group of 100 candidates mean =40. Later on it was discovered that the value 45 was
misread as 54. Find correct mean value.

Solution

= 40, n =100

=

40 =

=4000-54+45=3991

Correct mean is =

= 39.91

Example

The second of the two samples of 50 items is with mean 15. If the whole group of 150 items is
having mean of 16, find mean of first sample.

Solution

Combined mean =

& are of sample sizes of 1st & 2nd set

1 & 2 are mean

+ = 150

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Chapter 2 Mathematics

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30
th

Cross, 10
th

Main, Jayanagar 4
th

Block, Bangalore-11

= 50 2 = 15

= 100 1 =

16 =

1 = 16 .5

Difference between Binomial distribution and Poisson Distribution

- Poisson distribution is the limiting case of binomial distribution.

- When numbers of trial are large we use poison distribution and when number of trials are
small then we use Binomial distribution.

- In Poisson Distribution probability p is very small but not in Binomial distribution.

Normal Distribution

If mean is 0’0

Then the mean of the absolute value of normal random variable X = Variance √

- If for uniform distributed function in a region a x b

Variance is =
( )

And standard deviation: √Variance

= √
( )

=

 A regression model is used to estimate a value of y with a value of x, when y as a function
of another variable.

 Variance is used to determine whether variances in two or more, populations are
significantly different or not.