Concept of Probability and Probability Distribution

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The Concept of Probability and

The Concept of Probability and Probability Distribution

Probability Distribution

The concept of probability applies to a

The concept of probability applies to a random event random event . It is associated with chance and its. It is associated with chance and its prediction in mathematical terms.

prediction in mathematical terms.

Example: Example:

The outcome of a coin tossing, the throwing of dice in the game of Ludo, hitting a target The outcome of a coin tossing, the throwing of dice in the game of Ludo, hitting a target with a gun, whether it

with a gun, whether it will rain today etc.will rain today etc.

If a similar event occurs many times, sometimes the outcome o

If a similar event occurs many times, sometimes the outcome o f the events is favorablef the events is favorable (that means that I am looking for) and other times it is not. For example, if we toss a coin (that means that I am looking for) and other times it is not. For example, if we toss a coin the outcome will be either Head

the outcome will be either Head or Tor Tail. Suppose, I want Head, then ail. Suppose, I want Head, then one of the twoone of the two outcomes is favourable here. In ca

outcomes is favourable here. In case of Dice throwing, if I want six, then se of Dice throwing, if I want six, then one of the 6one of the 6 possible outcomes is favourable. Thus if the dice is face up with ‘six’, it is favourable possible outcomes is favourable. Thus if the dice is face up with ‘six’, it is favourable

otherwise it is not. otherwise it is not.

We can calculate the relative frequency of the favouravle events by taking the ratio of We can calculate the relative frequency of the favouravle events by taking the ratio of number of favourable events with the

number of favourable events with the total number of events. This ratio can be total number of events. This ratio can be different if different if the number of events or

the number of events or trials are different. When the total number of events (or trials) istrials are different. When the total number of events (or trials) is made very large, the relative frequency tends towards a

made very large, the relative frequency tends towards a fixed value. This can then be fixed value. This can then be saidsaid to be the probability of occurrence

to be the probability of occurrence of that event. It is a positive vof that event. It is a positive value, a fraction betweenalue, a fraction between 0 and 1 by definition.

0 and 1 by definition.

Definition of Probability: Definition of Probability:

Probability (p) is the ratio of the number of favourable

Probability (p) is the ratio of the number of favourable events (n) to the total number of events (n) to the total number of events when the total number

events when the total number of events (N) is made very large.of events (N) is made very large. N

N n n p

p

==

, where, where N N _{is very large.}_{is very large.}

Thus if

Thus if nn == N N , all the events are favourable,, all the events are favourable, p p

==

11..

When no events are favourable,

When no events are favourable, nn == 00,, p p

==

00. We have. We have00

<<

p p

<<

11.. Suppose,

Suppose, mm

==

the number of events that are the number of events that are not favourable. Wnot favourable. We can write,e can write, nn

++

mm

==

N N .. ∴

∴WWe can write, the probability of not e can write, the probability of not occurring the particular outcome isoccurring the particular outcome is N N n n N N n n N N N N m m q q

==

−−

==

11

−−

== 11

−−

p p.. Thus we have

Thus we have p p

++

qq

==

11.. This says, in any trial, something must happen, either in favouThis says, in any trial, something must happen, either in favou r r or not in

or not in favourfavour.. Example:

Example:

The probability of occurring Head in a coin toss is

The probability of occurring Head in a coin toss is p p

==

11//22_{. The probability of not}_{. The probability of not} occurring Head (or

occurring Head (or occurring Tailoccurring Tail) is) is qq

==

11//22.. ThusThus p p

==

qq

==

11//22

++

11//22

==

11. In other way. In other way we can say that the

we can say that the probability of occurring Headprobability of occurring Head

2 2 1 1

==

H H p

p and the probability of and the probability of occurring

occurring TTailail

2 2 1 1

==

T T p p , so that, so that p p_H_H

++

p p_T_T

==

11..

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Thus the sum of the probabilities of all the events is = 1. This is also called Thus the sum of the probabilities of all the events is = 1. This is also called Normalization.

Normalization.

For a dice throw, among 6 different outcomes, the probability of occurring one, is For a dice throw, among 6 different outcomes, the probability of occurring one, is

6 6 1 1 1 1

==

p p , the probability of occurrence of two is

, the probability of occurrence of two is

6 6 1 1 2 2

==

p

p and so on. Thusand so on. Thus

==

++

22 33 44 55 66 1 1 p p p p p p p p p p p p

++

6 6 1 1

++

6 6 1 1

++

6 6 1 1

++

6 6 1 1

++

6 6 1 1 6 6 1 1 =1.=1. Problem: Problem:

A box contains 4 identical balls, one red and

A box contains 4 identical balls, one red and three blue. What is the probability of three blue. What is the probability of drawing a red ball from the box?

drawing a red ball from the box?

Disjoint events: Disjoint events:

If two random events A and B do not occur at the same time, that is, either A occurs or B If two random events A and B do not occur at the same time, that is, either A occurs or B and one is not

and one is not dependent on other, they are called disjoint (mutually exclusive) events.dependent on other, they are called disjoint (mutually exclusive) events.

Addition of Probabilities: Addition of Probabilities:

Suppose in a given experiment, a random event A occurs with probability

Suppose in a given experiment, a random event A occurs with probability p p(( A A)) _and_and another with probability

another with probability p p(( B B)). If the two events are disjoint or mutuall. If the two events are disjoint or mutually exclusive, theny exclusive, then the joint probability of the two events, that

the joint probability of the two events, that is, the probability that either the event A or Bis, the probability that either the event A or B will take place, is given by

will take place, is given by

P

P (A or B) = P(A) + P(B).(A or B) = P(A) + P(B). Two events are

Two events are complimentarycomplimentary when no other events occur other than these. In that casewhen no other events occur other than these. In that case P(A) + P(B) = 1. In this case it is said that

P(A) + P(B) = 1. In this case it is said that the events A the events A and B form aand B form a complete groupcomplete group.. If the random events A, B, C, D form a complete group then P(A)+P(B)+P(C)+P(D)=1. If the random events A, B, C, D form a complete group then P(A)+P(B)+P(C)+P(D)=1.

Example: Example:

In the case of coin

In the case of coin tossing, P(H or T) = P(H) + P(T) = ½ + ½ =1.tossing, P(H or T) = P(H) + P(T) = ½ + ½ =1. In the case of Dice

In the case of Dice throwing, the probability of occurring ‘six’ or ‘one’,throwing, the probability of occurring ‘six’ or ‘one’, P(6 or 1) = P(6) + P(1) =1/6 + 1/6 =

P(6 or 1) = P(6) + P(1) =1/6 + 1/6 = 2/6.2/6.

Example:

Example: Two fair dice areTwo fair dice are thrown. What will be the probability that the total score isthrown. What will be the probability that the total score is either 10 or each score on each dice is more than 4?

either 10 or each score on each dice is more than 4?

The event A: (5,5), (4,6), (6,4)

The event A: (5,5), (4,6), (6,4)∴∴P(A) =P(A) =

12 12 1 1 36 36 3 3

==

. The event B: (5,5), (5,6), (6,5), . The event B: (5,5), (5,6), (6,5), (6,6)(6,6)

∴ ∴P(B) =P(B) = 9 9 1 1 36 36 4 4

==

∴∴P(A or B) =P(A or B) = 6 6 1 1 36 36 6 6

==

. However, P(A) + P(B) =. However, P(A) + P(B) =

36 36 7 7 9 9 1 1 12 12 1 1

==

++

. Thus. Thus P(A or B)

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Multiplication of Probabilities: Multiplication of Probabilities:

If the two events A and B are independent then the joint probability of occurring the two If the two events A and B are independent then the joint probability of occurring the two events at the same time is

events at the same time is

P

P (A and B) = P(A) + P(B).(A and B) = P(A) + P(B).

Example: Example:

If we toss two coins together, then there will be 4

If we toss two coins together, then there will be 4 kinds of events (HH, HT, TH, kinds of events (HH, HT, TH, TT). For TT). For any of the events to o

any of the events to occur, the probability isccur, the probability is P P _HH_HH

==

P P _HT_HT

==

P P _TH_TH

==

P P _TT_TT

==

11//44. Again, we. Again, we may see that

may see that P P _HH_HH

++

P P _HT_HT

++

P P _TH_TH

++

P P _TT_TT

==

11. . So we mSo we may say that ay say that the probability the probability of occurringof occurring either HH or HT or TH or TT is given by the rule of

either HH or HT or TH or TT is given by the rule of addition of probabilities.addition of probabilities.

Also we see that the probab

Also we see that the probability of any of the events, sayility of any of the events, say P P HH HH

==

××

==

P P H H

××

P P H H

2 2 1 1 2 2 1 1 4 4 1 1 and and T T H H HT HT P P P P P P

==

××

==

××

2 2 1 1 2 2 1 1 4 4 1 1

etc. Note that the event of head or tail in two coins happens etc. Note that the event of head or tail in two coins happens the same time and they are

the same time and they are independent. The joint probability of any number of independent. The joint probability of any number of independent events taking place

independent events taking place at the same time is the product of at the same time is the product of the probabilities for thethe probabilities for the individual events.

individual events.

Conditional Probability: Conditional Probability:

If A

If A and B are two eveand B are two events and the event B hnts and the event B happens only when A happens, the conditionalappens only when A happens, the conditional probability of B given A is denoted by P(B/A).

probability of B given A is denoted by P(B/A). ∴

∴The probability that both the events happen together isThe probability that both the events happen together is P(A and B) = P(B/A)

P(A and B) = P(B/A)

××

P(A).P(A).

Problem:

Problem: In a population of 100, there are 60 men and 40 women. Among the men 20In a population of 100, there are 60 men and 40 women. Among the men 20 are graduates and among

are graduates and among the women 10 are the women 10 are graduates. What is the probability of findinggraduates. What is the probability of finding a graduate woman if we pick a person randomly from the population?

a graduate woman if we pick a person randomly from the population? Solution.

Solution. _{If we choose from the women on}_{If we choose from the women only}_{ly, the probability that the woman will be}_{, the probability that the woman will be} graduate is P(G/W) = graduate is P(G/W) = 4 4 1 1 40 40 10 10

==

. . However, the However, the probability that the chosen person probability that the chosen person will be awill be a woman is P(W) = woman is P(W) = 10 10 4 4 100 100 40 40

==

. . Thus the probability Thus the probability that the chosen persthat the chosen person will be on will be womanwoman and a graduate is P(W and G) = P(G/W).P(W) =

and a graduate is P(W and G) = P(G/W).P(W) =

10 10 1 1 10 10 4 4 4 4 1 1

==

××

..

Find out the probability that the randomly chosen

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Probability of compatible events: Probability of compatible events:

Random events A and B are compatible if in a given trial or experiment both

Random events A and B are compatible if in a given trial or experiment both the eventsthe events can occur. The joint probability of two compatible events is calculated from the formula: can occur. The joint probability of two compatible events is calculated from the formula:

P(A or B) = P(A) + P(B) – P(A and B). P(A or B) = P(A) + P(B) – P(A and B).

Problem:

Problem:The probability of hitting a target by a gun A is 8/10 and by another gun B isThe probability of hitting a target by a gun A is 8/10 and by another gun B is 7/10. Two guns are fired together. What is the probability that the

7/10. Two guns are fired together. What is the probability that the target can betarget can be destroyed?

destroyed? Solution:

Solution: _{If any of the guns hits the target, it will be}_{If any of the guns hits the target, it will be destroyed. Al}_{destroyed. Also, both the guns}_{so, both the guns can}_can hit the target. Thus the probability of destroying the target is =

hit the target. Thus the probability of destroying the target is =

10 10 7 7 .. 10 10 8 8 10 10 7 7 10 10 8 8

−−

++

== 100 100 94 94 = = 0.94. 0.94.

How do we calculate average in

How do we calculate average in a probabilistic event?a probabilistic event?

Normally

Normally, we calculate the , we calculate the average (or mean) of a variable average (or mean) of a variable by summing up the valuesby summing up the values given and dividing up by the number :

given and dividing up by the number :

n n x x x x x x x x

==

11

++

22

++

...

++

nn .

. This is This is arithmetic arithmetic average.average.

In a probabilistic event, the ave

In a probabilistic event, the average is calculated in the following way:rage is calculated in the following way: ... ... 2 2 2 2 1 1 1 1

××

++

××

++

==

∑

x x P P x x P P x x P P x x _ii ii ii

For the average of square: For the average of square:

... ... 2 2 2 2 2 2 1 1 2 2 1 1 2 2 2 2

₌₌

_××

₊₊

_××

₊₊

∑

x x P P x x P P x x P P x x _ii ii ii

In the previous case of

In the previous case of arithmetic average it may be assumed that the arithmetic average it may be assumed that the events occur at aevents occur at a constant

constant probabilityprobability,,

n n P

P

==

11 . This means any of the events is equa. This means any of the events is equally probable.lly probable.

What is a probability distribution? What is a probability distribution?

A probability distribution is a set of probabilities describing the chan

A probability distribution is a set of probabilities describing the chan ces of events toces of events to occur.

occur. Any frequency distribution can be turned into an Any frequency distribution can be turned into an empirical probability distribution.empirical probability distribution. When we calculate the d

When we calculate the distribution from a finite sample, it is a frequency distribution.istribution from a finite sample, it is a frequency distribution. How is it calculated?

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Let us take some data. In order to obtain a distribution, we divide the data into different Let us take some data. In order to obtain a distribution, we divide the data into different classes or groups (like different age groups in a class or

classes or groups (like different age groups in a class or at a place). Wat a place). We check how mane check how manyy data fall into a class or group. The

data fall into a class or group. The ratio of the number in a ratio of the number in a class (class frequency) to theclass (class frequency) to the total number of data is called the

total number of data is called the relative frequencyrelative frequency. The relative frequency is assumed. The relative frequency is assumed to be the

to be the empirical probabilityempirical probability. W. We generally represent them e generally represent them graphically through agraphically through a bar bar diagram

diagram. This empirical probability tends towards actual probability. This empirical probability tends towards actual probability P P (( x x)) at some pointat some point x

x when we take the number of data to be very large.when we take the number of data to be very large. The

The distribution curvedistribution curve is a plot of is a plot of P P (( x x)) againstagainst x x wherewhere P P (( x x)) is the probability for theis the probability for the data point

data point x x to occur.to occur.

There are some well known probability distributions: (i) Normal or Gaussian distribution, There are some well known probability distributions: (i) Normal or Gaussian distribution, (ii) Binomial distribution, (iii) Poisson distribution.

(ii) Binomial distribution, (iii) Poisson distribution.

Normal or Gaussian distribution

Normal or Gaussian distribution is that which normally occurs for random events. Theis that which normally occurs for random events. The distribution curve looks like a bell shaped

distribution curve looks like a bell shaped curve, symmetrically spread around the meancurve, symmetrically spread around the mean value

value x x ..

The spread of the distribution is given by the

The spread of the distribution is given by the variancevariance 22 σ

σ (sigma-square) by the(sigma-square) by the

following formula: following formula:

( (

))

22 2 2 11

∑

−−

==

ii ii x x x x n n σ σ ₌₌ 22 22 x x x

x

−−

, for a sample of size, for a sample of size nn (( nn numbers of data points). Thenumbers of data points). The square root of

square root of 22 σ

σ is called theis called the standard deviation, SD =standard deviation, SD = σ σ ..

When the value of

When the value of σ σ 22 is small,is small, the spread of the distribution is small. When the value of the spread of the distribution is small. When the value of 2

2 σ

σ is large,is large, the spread of the distribution is also large. So, the variance the spread of the distribution is also large. So, the variance gives thegives the

dispersion of data points from the mean va dispersion of data points from the mean value.lue.

The mathematical formula for the above normal curve is The mathematical formula for the above normal curve is

( (

))

[ [

22 22

]]

2 2 // exp exp 2 2 1 1 )) (( σ σ σ σ π π x x x x x x P

P

==

−−

_,_{, where}_{where "exp" denotes}_{"exp" denotes the}_the exponential function,

exponential function, ee = 2.71828. The distribution is symmetric and the peak (most= 2.71828. The distribution is symmetric and the peak (most probable value) of a normal distribution occurs at

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Approximately 95% of area under curve falls within 2 SDs on either side of mean, and Approximately 95% of area under curve falls within 2 SDs on either side of mean, and

about 68% of the

about 68% of the area falls within 1 SD from mean. Tarea falls within 1 SD from mean. Total area under the curve otal area under the curve is 100%.is 100%.

Obtaining Normal Distribution: Obtaining Normal Distribution: Throwing of a dice.

Throwing of a dice.

A dice has 6 faces. As we throw the dice randomly any of the scores 1, 2,

A dice has 6 faces. As we throw the dice randomly any of the scores 1, 2, 3, 4, 5, 6 3, 4, 5, 6 willwill appear with equal probability

appear with equal probability P P

==

11//66. If we go on throwing the dice randomly and. If we go on throwing the dice randomly and indefinitely

indefinitely, we ma, we may generate a large ny generate a large number of scores. We may think that this set of umber of scores. We may think that this set of large number of scores is our population. Now we

large number of scores is our population. Now we want to take samples from thiswant to take samples from this population to calculate the properties of this population.

population to calculate the properties of this population.

Sample size 1:

Sample size 1: (taking one data at a time)(taking one data at a time) For a single throw of the dice,

For a single throw of the dice, the probability distribution is given by the following table.the probability distribution is given by the following table.

x 1 2 3 4 5 6

P

P((xx)) 11//66 11//66 11//66 11//66 11//66 11//66

The mean value of the distribution

The mean value of the distribution x x == 33..55 2 2 7 7 6 6 6 6 5 5 4 4 3 3 2 2 1 1

==

++

.. The variance, The variance, σ σ 22

==

∑

x x_ii22 P P _ii

−−

x x22 I I == 12 12 35 35 2 2 7 7 6 6 1 1 6 6 6 6 1 1 5 5 6 6 1 1 4 4 6 6 1 1 3 3 6 6 1 1 2 2 6 6 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2

₌₌













−−

××

++

××

++

××

++

××

++

××

.. Sample size 2:

Sample size 2: (taking two data at a time)(taking two data at a time)

)) (( x x P P x x x x Normal Distribution Normal Distribution

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If we throw two dice together, we shall obtain (1,1), (1,2),

If we throw two dice together, we shall obtain (1,1), (1,2), (2,1), (2,2) ….and so on. There(2,1), (2,2) ….and so on. There are 36 possible outcomes. We now consider the mean value of the two

are 36 possible outcomes. We now consider the mean value of the two data as our newdata as our new variable. The probability table is as follows.

variable. The probability table is as follows.

2 2 2 2 1 1 x x x x x x

==

++

11 11..55 22 22..55 33 33..55 44 44..55 55 55..55 66 P(x) P(x) 36 36 1 1 36 36 2 2 36 36 3 3 36 36 4 4 36 36 5 5 36 36 6 6 36 36 5 5 36 36 4 4 36 36 3 3 36 36 2 2 36 36 1 1 If we plot

If we plot P P (( x x)) _against_{against x}_{x , we get a}_{, we get a symmetrical distribution with the peak at}_{symmetrical distribution with the peak at} x x== 33..55.. W

We can ce can check thatheck that x x

==

₃₃_..₅₅ from the symmetry of the from the symmetry of the distribution.distribution.

The variance, The variance, 24 24 35 35 )) 5 5 .. 3 3 (( ...} ...} 36 36 2 2 )) 5 5 .. 1 1 (( 36 36 1 1 1 1 { { 22 22 22 2 2

₌₌

_××

₊₊

_××

₊₊

₋₋

₌₌

σ σ ..

In this way if we check

In this way if we check with sample size 3 (that means, throwing 3 dice with sample size 3 (that means, throwing 3 dice together) andtogether) and more, we obtain a c

more, we obtain a clear symmetric distribution with mean value atlear symmetric distribution with mean value at x x

==

33..55. The spread. The spread of the distribution (variance) becomes consistently smaller as we go on

of the distribution (variance) becomes consistently smaller as we go on taking larger taking larger samples. The distribution takes a definite symmetrical shape around the mean

samples. The distribution takes a definite symmetrical shape around the mean value for value for large sample sizes. This distribution is

large sample sizes. This distribution is normal or Gaussian distributionnormal or Gaussian distribution..

⇑⇑

This lecture note is compiled from many sources (Books and This lecture note is compiled from many sources (Books and websites) as anwebsites) as an

introduction to the subject. This is intended to be distributed privately among students. introduction to the subject. This is intended to be distributed privately among students.

Books:

Books: 1. Statistics 1 & 2 (Advanced Leve1. Statistics 1 & 2 (Advanced Level Mathematics), Steve Dobbs and Jane Miller.l Mathematics), Steve Dobbs and Jane Miller. Cambridge University Press. 2. Differential and Integral

Cambridge University Press. 2. Differential and Integral Calculus –II , N. Calculus –II , N. PiskunovPiskunov, MIR , MIR pub., Moscow some websites.

pub., Moscow some websites.

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