3 Chapter 2
Mathematics
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 alike and are alike =
!
! !
are alike,
are
!
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
!
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 38
3 Chapter 2
2.
Mathematics
Combination Number of combinations of n different objects taken ‘r’ at a time (without repetitions) ! is nCr = ! !
Number of combinations of n different objects taken ‘r’ at a time (with repetitions) is given by n+r-1Cr
Example Given 3 letter A, B, C→ select 2 at a time Permutations (without repetitions) → AB, BC, CA, BA, CB, AC (Order does matter) Permutations (with repetitions) → AB, BC, CA, BA, CB, AC, AA, BB, CC Combinations (without repetitions) → AB, BC, CA (Order does not matter) Combinations (with repetitions) → AB, BC, CA, AA, BB, CC Properties of combinations a) b)
nCr
= nCn-r nCr = n-1Cr + n-1Cr-1 (Pascal's formula)
Binomial coefficient The number nCr or C(n,r) is called a binomial coefficient and this is written as ( ) 3.
Event Outcome of an experiment is called event.
4.
Exhaustive events (Sample space) A set of events is said to be exhaustive if it includes all possible events (i.e. the set of all permutations and combinations of an experiment).
5.
Mutually exclusive events (Disjoint events) Two events are called mutually exclusive, if the occurrence of one excludes the occurrence of others i.e. both can’t occur simultaneously. =φ, P(
A
) =0
B
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 39
3 Chapter 2
6.
Mathematics
Equally likely events If one of the events cannot happen in preference to other, then such events are said to be equally likely.
Note Throwing a die (also turning up of the six different faces of the die) is exhaustive, mutually exclusive and equally likely events. Odds in favour of an event = Where m→ no. of ways favourable to A n→ no. of ways not favourable to A Odds against the event = 7.
Probability If there are ‘n’ exhaustive, mutually exclusive and equally likely cases of which ‘m’ are favourable to an event A, then probability of the happening of A is P(A)=
=
. .
P(A)+ P(A’)=1 Important point
P(A B)→ Probability of happening “at least one” event of A & B P(A B) )→ Probability of happening “both” events of A & B If the events are certain to happen, then the probability is unity. If the events are impossible to happen, then the probability is zero.
8.
Addition law of probability a. For every events A, B and C not mutually exclusive P(A B C)= P(A)+ P(B)+ P(C)- P(A B)- P(B C)- P(C A)+ P(A B C) b. For the event A, B and C which are mutually exclusive P(A B C)= P(A)+ P(B)+ P(C)
9.
Independent events Two events are said to be independent, if the occurrence of one does not affect the occurrence of the other. If P(A B)= P(A) P(B)
10.
Independent events A & B
Conditional probability If A and B are dependent events, then P( ) denotes the probability of occurrence of B when A has already occurred. This is known as conditional probability. P(B/A)=
(
) ( )
For independent events A & B → P(B/A) = P(B) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 40
3 Chapter 2
11.
Mathematics
Theorem of combined probability If the probability of an event A happening as a result of trial is P(A). Probability of an event B happening as a result of trial after A has happened is P(B/A) then the probability of both the events A and B happening is P (A B)= P(A). P(B/A) ,
[ P(A) 0]
= P (B). P(A/B) , [ P(B) 0] This is also known as multiplication theorem. For independent events A&B → P(B/A) = P(B), P(A/B )= P(A) Hence, P(A B) = P(A) P(B) Important point If
&
1. 2. 3. 4.
(1- ) → probability of first event happens and second fails (i.e only first happens) (1- )(1- ) → probability of both event fails 1-(1- )(1- ) → probability of at least one event occur → probability of both event occurs
12.
are probabilities of two independent events then
Baye’s Theorem An event A corresponds to a number of exhaustive events
,
,..,
.
If P( ) and P(A/ ) are given then, (
P( )=
(
). ( ) ). ( )
This is also known as theorem of Inverse Probability. Example A professor have 3 research students A, B, C. He assigns 40% of the problem to A →of which A solves 3% incorrectly He assigns 25% of the problem to B →of which B solves 5% incorrectly He assigns 35% of the problem to C →of which C solves 4% incorrectly What is probability that a randomly selected, wrongly solved problem was that assignment to A? Solution Let
A= Event of assigning the problem to A B= Event of assigning the problem to B C= Event of assigning the problem to C P (A) =
,
P (B) =
,
P(C) =
A
B
C
=
W=the event of problem solved incorrectly P (W/A) =
,
P (W/B) =
,
P (W/C) =
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 41
3 Chapter 2
Mathematics
P (A/W) =? ( ) ( )
P (A/W) = =
( ) ( )
( ) ( )
. .
.
.
( ) ( )
= 0.31
Example A box contains 10 white balls and 8 black balls. Calculate the probability of drawing 2 White and 1 Black balls for the following three cases: a. All the 3 balls picked in a single draw b. All the 3 balls drawn one after another (replacement) c. All the 3 balls picked one after another (without replacement) Solution a. All the ball in a single draw =
=
b. One after another (replacement) = c. One after another (without replacement) =
= =
= =
=
Example 3 of 6 vertices of a regular hexagon are chosen at random. What is the probability that the chosen triangle is equilateral? 1
2
6
3
5
4
Solution total=
!
= !=
=20
Option with equilateral ∆=2
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 42
3 Chapter 2
Probability of the chosen triangle to be equilateral=
Mathematics
=
Example A problem is given to students A, B, C whose chances of solving it is , , .What is probability that problem will be solved? Solution Approach 1 Probability that A solves the problem P (A) = , P (A’) = P (B) =
, P (B’) =
P(C) =
, P(C’) =
Probability that A, B, C can’t solve the problem = (1
) (1
) (1
)= x x =
Probability that problem will be solved=1 -
=
Approach 2 (Lengthy, Not recommended) Probability that problem will be solved = ABC+ ABC’+ AB’C+ AB’C’+ A’BC+ A’BC’+ A’B’C =
+
= (
+
+
+ +
+ +
+
+ +
+
+
)=
Example In a race, the odds is favour of four persons probability that one of them wins the race.
,
,
,
are 1:3, 1:4, 1:5, 1:6 respectively. Find
Solution = ,
= ,
= ,
=
Since all the events are mutually exclusive Probability that one of them wins= + + + =
Example (Hungarian problem) A & B throw alternatively with a pair of dice.’ A ‘ wins if he throws ‘ 6 ‘before ‘ B ‘throw ‘ 7 ‘ .’ B ‘wins if he throws ‘ 7 ‘before ‘ A ‘throws ‘ 6 ‘. If ‘ A ’ starts the game, find probability of winning ‘A ‘ Solution →6
(3 ,3 ) ( 4, 2 ) ( 2 , 4 ) ( 1 ,5 ) ( 5 ,1 ) =
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 43
3 Chapter 2
B→7
( 3,4 ) ( 4 ,3 )( 2 ,5 ) ( 5 ,2 ) ( 1, 6 ) ( 6 ,1 ) =
P (A) =
, P( ) =
P (A) for winning = =
(1 +
, P(B) = +
+(
x
=
, P( ) = x
+
x
x
x
x
+ …………………
) + ……………………….)
{ We know that 1 + +
+ …………………….=
( r < 1 )}
{ since 36 36 – 31 30 = 6 (216 -155 ) = 6 ( 61 )}
=
13.
Mathematics
=
Repeated trials Under repeated trials condition, probability of ‘r’ successes in ‘n’ trials is Where, n = the total number of trials, p = probability of success, q = 1-p (probability of failure) Probability of at least ‘r’ success in ‘n’ trials = sum of probability of r, r+1,…..n success =
+
+……+
Example: Probability of coming exactly 4 heads, when a coin is tossed six times = ( ) ( ) 14.
Random variable Real variable associated with the outcome of a random experiment is called a random variable. It is process of assigning a number or numerical value to the outcome of the experiment.
Discrete random variable If a random variable takes a finite set of values, it is called a discrete random variable.
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 44
3 Chapter 2
Mathematics
Continuous random variable If a random variable takes infinite number of uncountable values, it is called a continuous random variable. Mixed random variable In this, some values may be discrete and some values may be continuous. All the random variables are described by: 15.
Probability Density Function (Discrete pdf) Cumulative Distribution Function (cdf) Discrete Probability Density Function (pdf) or Probability Mass Function The set of values Xi with their probabilities constitute a probability distribution or probability density function of the discrete variable X. If f(x) is the pdf, then f( ) = ( = ) , f(x)
16.
( )=1 0
Discrete Cumulative Distribution Function (cdf) or Distribution Function The cumulative distribution function F(x) of the discrete variable x is defined by, ( ) = F(x) = P(X x) =
( )=
( )
Note: For a discrete r.v, the CDF is staircase waveform. 17.
Continuous probability density function (pdf) When a variable X takes every value in an interval, it gives rise to continuous distribution of X. Probability distribution function of a continuous variable X is defined by a function f(x) such that the probability of the variable x falling in the small interval x - dx to x + dx is f(x)dx. Then f(x) is called the probability density function. f(x) = 0
xb
b
Range may be finite or infinite
PDF has the following properties 1. 2.
Probability Density function is always positive ( ) =1 ∫ THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 45
3 Chapter 2
18.
Mathematics
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 ( )
( ) =f(x) 0 i) = ( ) 0 ii) 1 iii) If ( ) ( ) In other words, CDF is monotone (non-decreasing function) ( ) =0 iv) ( ) =1 v) ( ) -∫ ( ) = ( ) 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
P (Xi) k (A) k =?
1
2
3
4
5
6
3k
5k
7k
9k
11k
13k
(B) P [ x < 4 ] =?
( C ) P [3 < x
6 ] =?
Solution ( )=1
(A)
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) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 46
3 Chapter 2
Mathematics
= 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 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 1 5/6 4/6 3/6 2/6 1/6
1
2
3
Fx(x) = u(x+1)+ u(x+2)+
4
5
(x+3)+
6
x
(x+4)+ u(x+5)+ u(x+6)
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
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 47
3 Chapter 2
a. b.
Mathematics
P[ x a ] = P[ x a ] P [ x > b ] = 0.05
Solution a.
( )
∫
=∫
( )
⇒∫ 3 =∫ 3 3 3 ⇒ = 1-α ⇒ 2 3=1 ⇒ = √
b.
P ( x> b ) = 0.05 ∫ 3 3.
(
= 0.05 )
= 0.05
= 0.95 ⇒ b = ( ) 19.
Expectation E(x) The mean value of the probability distribution of a variety is commonly known as its expectation. E(X) = E(X) = ∫
f( ) (Discrete case) f( )
(Continuous case)
Properties of expectation 1. 2. 3. 4.
E(constant) = constant E(CX) = C . E(X) [C is constant] E(Ax+By) = A E(x)+B E(y) [A& B are constants] E(XY)= E(X) E(Y/X)= E(Y) E(X/Y) E(XY) E(X) E(Y) in general But E(XY) = E(X) E(Y) , if X & Y are independent
20.
Variance Var(X)
Var X =E (x
) ]
Var X= (
) f( )
Var X=∫ (
(Discrete case)
) f( )dx (Continuous case)
It can be proved that E(
)- E(X)
Properties of Variance 1. Var(constant) = 0 2. Var(Cx) = C Var(x) -Variance is non-linear [Here C is constant] 3. Var(Cx D)= C Var(x) -Variance is translational invariant [C&D are constants] THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 48
3 Chapter 2
4. Var(x-k) = Var(x) 5. Var(ax+by) =
Mathematics
-[k is constant] Var(x) +
Var(y)
2ab cov(x,y) (if not independent)
[A & B are constants] = 21.
Var(x) +
Var(y) (if independent)
Covariance Cov (x,y)=E(xy)-E(x) E(y) If independent
covariance=0,
E(xy) = E(x) . E(y)
(if covariance=0, then the events are not necessarily independent) Properties of covariance 1. 2. 3. 22.
23.
Cov(x,y) = Cov(y,x) (i.e. symmetric) Cov(x,x) = Var(x) |Cov(x,y)| Standard distribution function (Discrete r.v. case) 1.
Binomial distribution { P(x=r) =
2.
Poisson distribution P ( =
}
)=
!
Standard distribution function (Continuous r.v. case) (
1.
Normal distribution : { f(x) =
2.
Exponential distribution : { f(x) =
}
√
,
= 0, 3.
Uniform distribution: { f(x)=
24.
0 ,
- x1) = P(x < x1) P( x < x1) = 0.5 + P(0 < x < x1)
Example In a class, if on an average one student in every ten is having swine flu, find the probability that out of 5 students expected to attend a class, at least 4 will not have swine flu? Solution P (student not having swine flu) = P (4 students safe) = =
( ) ( )
pr qn-r
=5
P (5 students safe) =
( ) . ( ) =1
Total= P(4 students safe)+P(5 students safe) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 51
3 Chapter 2
=5
+
=
=( )
Mathematics
.
Example Probability of an individual coal miner being killed during a year is . Calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year. Solution Approach-1 P=
,
n = 200 𝛌 = np =
=
P(atleast one fatal) = P(1) +P(2) +P(3) +……+P(200) = 1-P(0) = 1 =1
!
= 1-0.92=0.08
Approach-2 The same problem can be solved by Binomial distribution method: P (atleast one fatal) = 1 =1
(
(
) (
)
)
= 1- 0.92=0.08 Example If 10% of cars produced in a car factory is defective. Determine the probability that out of 10 cars chosen at random (i)1 (ii) None (iii) At most 2 , cars will be defective. Solution P (defective) = P (not defective) =
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 52
3 Chapter 2
(i) P(1 defective) =
Mathematics
( ) ( )
(ii) P(none defective) =
( ) ( )
(ii) P(at most 2 defective) =P (0 defective) +P (1 defective) +P (2 defective) P (2 defective) =
( ) ( )
Example It is known that Students of a class score marks(which is normally distributed) with a mean of marks=60 and S.D=5. What % of students score more than 60 marks? f(z)
0 Solution x=60, =60, z=
=
z
=5 =0
i.e, if x>60, z>0 and hence the area lying to the right of z=0 is 0.5. 50% students score more than 60 marks. Example A manufacturer produces resistor with resistance between 98 and 102 ohms?
=100Ω and
=2Ω. What % of resistor will have
f(z)
z
z
z
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 53
3 Chapter 2
Mathematics
Solution z2=
=
=1
z1=
=
= -1 (
Area between =∫ 27.
and
=∫
)
dx
√
=2 0.3473 =0.6828
√
Measure of Central Tendency It gives central idea of whole data. It is a representative of a set of data. Measure of central tendencies: a) Mean b) Median c) Mode
a. Mean For a set of n values of a variant X=( The arithmetic mean, ̅ = .
,…..,
).
For a grouped data if , , … . . , are mid values of the class intervals having frequencies , ,….., ,then, ̅ = . If ̅ is mean for data; ̅ is mean for data; then Combined mean of
,
+
is taken ̅ =
̅
̅
.
If , be mean and SD of a sample size and , be those for a sample of size then SD of combined sample of size + is given by, (
+ =
(
)
=
-m )
= (
(m ,
+
+
+
= mean, SD of combined sample) )+ (
)
b. Median When the values in a data sample are arranged in descending order or ascending order of magnitude the median is the middle term if the no. of sample is odd and is the mean of two middle terms if the number is even. c. Mode It is defined as the value in the sampled data that occurs most frequently.
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 54
3 Chapter 2
Mathematics
Important points Mean is best measurement ( all observations taken). Mode is worst measurement ( only maximum frequency is taken). In median, 50 % observation is taken. Sum of the derivation about “mean” is zero. Sum of the absolute derivations about “median” is minimum. Sum of the square of the derivations about “mean” is minimum. 28.
Co-efficient of variation Ratio of S.D. and mean is called as co-efficient of variation. Coefficient of variation = ̅
29.
100
Correlation coefficient The numerical measurement of the correlation is called the correlation coefficient and is ( , ) defined by the relation, (x,y) =
Note 30.
-1 (x, y) 1 (x,y) = (y,x) |(x,y)| = 1 when P(x=0)=1; or P(x=ay)=1 [ for some a] If the correlation coefficient is -ve, then two events are negatively correlated. If the correlation coefficient is zero, then two events are uncorrelated. If the correlation coefficient is +ve, then two events are positively correlated.
Line of Regression If the scatter diagram indicates some relationship between two variables X &Y, then the dots of the scatter diagram will concentrate round a curve. This curve is called the curve of regression. When the curve is straight line it is called line of regression. The equation of the line of regression of y on x is
=
The equation of the line of regression of x on y is (
)=
̅̅̅̅
̅̅̅̅
31.
̅̅̅̅
(
̅̅̅̅
) (
)
is called the regression coefficient of y on x and is denoted by byx. is called the regression coefficient of x on y and is denoted by bxy.
Joint Probability distribution If X & Y are two random variable then
Joint distribution is defined as,
Fxy(x,y) = P(X
x;Y
y)
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 55
3 Chapter 2
Mathematics
Properties of Joint distribution function/ cummulative distribution function: 1. 2. 3. 4. 5.
F F F F F
32.
( , ) = 0 ( , ) = 1 ( , ) = 0 { F ( (x, ) = P(X x Y ( , y) = F (y)
,
) = P(X Y ) = F (x) . 1 = F (x)
y) = 0 x 1 = 0 }
Joint probability density function Defined as f(x, y) = Property: ∫
∫
F(x, y) 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 THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 56
3 Chapter 2
= 50
Mathematics
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
-
Variance is =
(
x b
)
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.
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 57
