Statistical Probability

## Probability and Statistics

In this Part, we discuss about Probability, Expected value and related topics like :

• Random Experiment
• Sample Space, Events & Outcome
• Statistical Probability Theories
• Total Probability Theorems
• Mutually Exclusive & Dependent Events
• Compound Probability
• Conditional Probability
• Permutation & Combination
• Bernoulli’s Theorem
• Expected Value

## Statistical Probability

In general terms, Statistical Probability refers to the chances of occurrence or non-occurrence of an event. In statistics, Probability is the measure of the likeliness (between the value 1 and 0) of occurrence of an event.

Terms related with Probability

• Experiment : Experiment is the situation involving chance or probability that leads to results, called outcomes.
• Outcome :  Outcome is the Result of a single trial of an experiment.
• Event  : Event is One or more outcomes of an experiment.
• Probability :  Probability is Measure of how likely an event is. Probability of occurrence of an event ‘A’ is defined as (known as Classic Definition of Probability):

P (A) = (Number of ways an event A event may occur) / Total Number of Possible Outcomes

Random Experiment

An experiment is called a random experiment if when conducted repeatedly under homogenous conditions, it gives different result with one of the various possible outcomes.

Ex. : If a dice is rolled out, there are 6 possible outcomes (1,2,3,4,5,6). So any number may be the outcome.

Sample space

The set of all possible outcomes of a random experiment is called sample space. It is denoted by letter ‘S’.

Ex: If 2 coins are tossed the set of all these outcome constitutes a sample space, expressed as

S = {HH, TT, HT, TH}

Statistical Definition of Probability

If the experiment is repeated a large number of times under necessary identical conditions, the limiting value of the ratio of the number of times the event A happens to the total number of trials of the experiment, as the number of trial increases indefinitely, is called the probabilities of occurrence of the event A.

Modern Definition of Probability

P(A) is called the probability of the class of events A, in a sample space S, where P is the probability measure (a real valued function defined on A), satisfying the following axioms

(i) 0£P(A)£1 for event belonging to å, (ii) P (S) = 1, (iii) For every finite or infinite sequence of disjoint events A1, A2, … An, P (A1 È A2È …) = P(A1) + P (A2) + …

## Statistics of Events

An Statistical Event is a Sub set of possible outcome in a sample space

Types of Events

• Sure event: An event which is sure to happen is called sure event. Probability of sure events become one.
• Impossible event: On the performance of a random experiment, an event which is impossible to happen.
• Simple event: On the performance of a random experiment, an event with a single possible outcome
• Elementary event: An element of a sample space, which shows all possible outcome of a random experiment
• Compound event: On the performance of a random experiment an event with two or more simple events
• Independent event: Two or more events are said to be independent, if they are not affected by each other
• Mutually Exclusive events: Two or more events are said to be mutually exclusive events, if more than one event is not possible at a time.
• Equally likely events: Two or more events are said to be equally likely if each of them has equal chance of occurrence or non-occurrence in preference to others.
• Complementary events: An event which consists in the negation of another event is called complementary event of the latter event.

## Statistical Probability Theorems

The Principal Theorems of outcome Statistical Events may be enumerated as follows.

• Events  Mutually Exclusive: For any two mutually exclusive events A and B, the probability that either A or B occurs, is given by the sum of individual probabilities of A and B.

This is expressed as : P(A È B) = P(A) + P(B)

• Events not Mutually Exclusive : If A and B are any two events, then

P(A or B) = P(A) + P(B) – P(A and B), expressed as : P(A È B) = P(A) + P(B) – P(A Ç B)

• Occurrence of any one event : For any three events A, B and C, the probability that at least one of the events occurs  is expressed as :

P (A È B È C) = P(A) + P(B) + P(C) – P(A Ç B) – P (A ÇC) – P(B Ç C) + P (A Ç B Ç C)

## Statistical Probability Theorems – Problems

Statistical Probability Theorems – Problems

Ex. 1 : A coin is tossed thrice. What is the probability of getting 2 or more heads?

If a coin is tossed three times, then we have the following sample space.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.  2 or more heads imply 2 or 3 heads.

If A and B denote the events of occurrence of 2 and 3 heads respectively, then we find that

A = {HHT, HTH, THH} and B = {HHH}. So, P(A) = n(A) / n(S) = 3/8,  So, P(B) = n(B) / n(S) = 1/8,

As A and B are mutually exclusive, the probability of getting 2 or more heads is

P(A È B) = P(A) + P(B) = (3/8) + (1/8) = 4/8 = 1/2, (or .50)

Ex. 2: The probability that an Accountant’s job applicant has a CWA Degree is 0.75, that he is a CA is 0.35 and that he is both CWA and CA is 0.20. Out of 600 applicants, how many would be CWA or CA?

Let the applicant is a CWA be denoted by B and that he is a CA be denoted by C

So, P(B) = 0.75, P(C) = 0.35 and P(B Ç C) = 0.20

The probability than an applicant is CWA or CA is given by

P(B È C) = P(B) + P(C) – P(B Ç C)

= 0.75 + 0.35 – 0.20

= 0.90

So, Number of candidates who are either CWA or CA is (.90 x 600) = 540

## Total Probability Theorems – Problems

Total Probability Theorems – Problems

Ex. 1 : Mutually Exclusive  Events

The probability that a company manager will travel by train is 2/3 & and that he will travel by Bus is 1/4. What is the probability of not travelling by Bus or Train

So, the probability of his traveling by train or bus is  P(Bus) + P (train) = (2/3) + (1/4) = 11/12

The probability of not travelling by Bus or Train is 1 – (11/12) = 1/12

Ex 2 : Events not Mutually Exclusive

There are three persons A, B and C having different ages. The probability that A survives another 6 years is 0.75, B survives another 6 years is 0.55 and C survives another 6 years is 0.45. The probabilities that A and B survive another 6 years is 0.40, B and C survive another 6 years is 0.30 and A and C survive another 6 years 0.40. The probability that all these three persons survive another 6 years is 0.20. Find the probability that at least one of them survives another 6 years.

Here, P(A) = 0.75, P(B) = 0.55, P(C) = 0.45,

P(A Ç B) = 0.40, P(B Ç C) = 0.30, P(A Ç C) = 0.40, P(A Ç B ÇC) = 0.20

The probability that at least one of them survives another 6 years is given by

P (A È B È C)

= P(A) + P(B) + P(C) – P(A Ç B) – P(A Ç C) – P(B Ç C) + P(A Ç B Ç C)

= 0.75 + 0.55 + 0.45 – 0.40 – 0.30 – 0.40 + 0.20 = .85

## Total Probability Theorem – Problems

Total Probability Theorems – Problems

Ex. 1 : In drawing one card from a deck of 52 cards, find the probability that a single draw will be either a face card or a spade card.

Let A = the event of drawing a face card. There are three face cards (jack, queen, king) in

each of the four suits (heart, spade, diamond and club).

Then A = {Jh, Qh, Kh, Js, Qs, Ks, Jd, Qd, Kd, Jc, Qc, Kc}

Let B = The event of drawing a spade card.

Then B = (Js, Qs, Ks, As, 2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s, 10s)

Notice that elements Js, Qs, Ks, which are elements of set B are also elements of set A, or set A and B = (Js, Qs, Ks),

Hence, P(A or B) = P(A) + P(B) – P(A and B) = (12/52) + (13/52) – (3/52) = 22/52 = 11/26

Ex. 2 : The probability that a contractor will get a plumbing contract is 3/4 and the probability that he will not get building contract is 4/5.  If the probability of getting at least one contract is 2/5, What is the probability that he will get both ?

Let A and B denote the events that’s the contractor will get plumbing and building contracts respectively. Then

P (A) = 3/4, P (`B) = 4/5, P (B) = 1-  P (`B) = 1-(4/5) = 1/5

Probability that contractor gets at least one contract = P(A È B) = 2/5

So, P(A) + P(B) – P(A Ç B) = 2/5, or, (3/4) + (1/5) + P(A Ç B) = 2/5

Or P(A Ç B) = (3/4) + (1/5) – (2/5) = 11/20

So probability that the contractor will get both the contracts is 11/20

## Compound Probability Theorem

Compound Probability Theorem enumerates Independent and Dependent Events

Independent Events

Probability that both Independent events. A and B will occur is P(A Ç B) = P(A) . P(B)

So, P(A and B) = P(A) . P(B)

Dependent Events : If events A and B are so related that the occurrence of B is affected by the occurrence of A, then A and B are called dependent events.

The probability of event B depending on the occurrence of event A is called conditional probability and is written as P (B ï A), which may be read, “the probability of B given A.”

The probability that both the dependent events A and B will happen is given by :

P(A Ç B) = P(A) . P(B ï A) or P(A and B) = P(A). P(B ï A)

## Compound Probability Theorem – Problems

Compound Probability Theorem- Problems

Ex 1 : Independent Events

A bag contains 3 red and 5 blue balls. One ball is drawn from the bag and then replaced. Another ball is drawn after the replacement. Find probability that both drawings are of red balls.

The occurrence on the first draw certainly has no relationship with the second draw since the second ball is drawn after the first ball is replaced. The two draws are independent.

Let  A = the event of the first draw that will have a red ball

Let B = the event of the second draw that will also have red ball.

Since there are 3 red balls in the bag of eight balls, in each draw we have P(A) = P(B) = (3/8)

So, P(A and B) = P(A) . P(B)   = (3/8) X (3/8) = 9/64

Ex. 2 : Dependent Events

A bag contains 3 red and 5 blue balls. One ball is drawn from the bag and is not returned to the bag when the second ball is drawn. Find the probability that both balls of the two drawings are red.

Here P(A) = 3/8 (as there are 3 red Ballsin the bag of eight balls)

The event B depends on event A, as when the first ball drawn is red the Probability for the second draw to have a red ball is p (B|A) = 2/7 (as there are 2 red balls in 7 balls, after the first red ball is drawn).

The probability that both balls in the two draws are red is

P(A and B) = P(A) . P(B ï A) = 3/8 X 2/7 = 6/56=3/28

## Compound Probability Theorem – Problems

Compound Probability Theorem- Problems

Ex 1 : In a group of 25 males and 15 females, 15 males and 8 females are Businessmen. What is the probability that a person selected at random from the group is a businessman, given that the selected person is a male?

Let B and M stand for businessman and Male respectively. We are to evaluate P(B/M).

(B Ç M) represents the event of both businessman and male.

P(B/M) = [{ P(B Ç M)} / { P(M)}] = [(15/40) / (25/40)] = 0.60

So, the  probability that a selected male is businessman is 0.60

Ex. 2 : Mr. Banerjee is selected for three separate posts. For the first post, there are four candidates, for the second, there are six candidates and for the third, there are 12 candidates. What is the probability that Mr. Banerjee would be selected?

Let the three posts be denoted by A, B and C respectively.

So, P(A) = 1/4, P(B) = 1/6, P(C) = 1/12

The probability that Mr. Banerjee would be selected (i.e. selected for at least one post)

= P(A È B È C) = 1 – [P(A È B È C)’] = 1 – P(A’ Ç B’ Ç C’) [(As A, B and C independent, so are their complements)]

= 1 – [ P(A’) x P(B’) x P(C’)]  ( By De Morgan’s Law).

= 1- [(1- 1/4) x (1- 1/6) x (1- 1/12)] = 1- {(3/4) x (5/6) x (11/12)} = 1- (55/96) = 41/96

## Theorem of Compound Probability – Problems

Compound Probability Theorem- Problems

Ex 1 : An article contains of two parts A and B. The production process of each part is such that probability of defect A is 0.18 and that in B is 0.15. What is the probability that the assembled product will not have any defect?

Let X be the event that part A is free from defect.

So, P(X) = 1 – P (Part A is defective) = 1 – 0.18 = 0.82

Let Y be the event that part B is free from defect

P(Y) = 1 – P (Part B is defective) = 1 – 0.15 = 0.85

The two events X and Y are independent, as part A having no defects or otherwise does not influence on part B’s being defective or otherwise. Let E be the event that the assembled product will not have any defect, i. e. when A and B free are from defects,

So, P(E) = P(X Ç Y) = P(X) . P(Y) ( as XY are independent events)

= 0.82 x 0.85 = 0.697

Ex 2: Two cards are drawn from a pack of cards in succession with replacement. Find the probability that both are Aces.

There are 4 Aces in a pack of 52 Cards. So,Probability of first card being an ace P(A) = 4/52 = 1/13

Probability of second card being an ace P(B) = 4/52 (when first card is replaced) = 1/13.

Both these events are independent.

So,  P(A x B) =P(A) x P(B) =(1/13) x (1/13) = 1/ 169

## Conditional Probability

General multiplication theorem is based on the assumption that each event is independent, i.e., the happenings of first event does not affect the second or subsequent events and in each trial conditions are identical.

However, Conditional Probability refers to the chances of outcome occurs when another event has also occurred. It may be stated as the probability of B given A and is written as P(B|A), where the probability of B depends on that of A happening.

Theorem of Conditional Probability

The probability of simultaneous occurrence of two dependent events is the product of the probability of the first event and the probability of the second after the first sub-event has occurred.

P(A & B) = P(A) x P(B/A), P(A and B) = P(B) x P(A/B), where P(B/A) means the conditional probability of B when A has already happened and  P(A/B) means the conditional probability of A when B has already happened.

Note: P(A & B) can also be written as P(A B)

So, for events A,B and C, the formula of conditional probability will be:

P(A  B and C) = P(A) x P(B/A) x P(C/A B), where P(A) = Probability of occurrence of A, P(B/A) = Probability of B, when A has already happened, P(C/A B) = Probability of C, when A and B both have occurred

Ex. Four cards are drawn one by one from a pack. We are to find the probability of the card being drawn is King.

The probability of first card being King is 4/52 or 1/13.  If the card drawn each time is not replaced, then each succeeding event will become dependent on previous event or event and the probability computed in such cases is called as Conditional Probability. It is also known as probability due to partial exhaustion of a sample.

## Conditional Probability – Problems

Conditional Probability – Problems

A bag contains 3 green and 7 blue balls. Two draw of a ball is made without replacement. What is the probability that both the balls are (i) green (ii) blue, (iii) of the same colour (iv) of different colours.

1. Probability of green ball in first draw = P(A) = 3/10(as there are 3 green balls out of 10 balls)

Probability of green ball in second draw after getting green ball in first draw = 2/9 (as there are 2 green balls out of 9 balls)

Probability that both the balls are green = (3/10) x (2/9) = 6/90 = 1/15

2. Probability of blue ball in first draw  = P(A) = 7/10(as there are 7 Blue balls out of 10 balls)

Probability of Blue ball in second draw after getting Blue ball in first draw = 6/9 (as there are 6 Blue balls out of 9 balls)

Probability that both the balls are Blue = (7/10) x (6/9) = 42/90 = 14/30

3. Both balls of the same colour means either both balls are green or both are blue

Probability that both balls are green = 1/15.  Probability that both balls are blue = 14/30

As, both these events are mutually exclusive, probability that both the balls are of the same colour = (1/15) + (14/30) = 16/30 = 8/15

4. Both balls of different colour means that the first is green second blue or first is blue and second green

Probability that first ball is green and second is blue

P(AB) = P(A) x P(B/A) = (3/10) X (5/9) = 15/90 = 1/6

Probability that first ball is Blue and second is Green = (7/10) x (3/9) = 21/90 = 7/30

Both these events are mutually exclusive. So, the probability of both balls being of different colurs is 1/6 + 7/30 = 12 / 30 = 2/5

## Statistical Permutation and Combination

Statistical Permutation and Combination – Problems

Permutations and combinations is mathematical theorem of computing  number of ways of objects from a set.

If a certain operation can be performed in any one of “n1” ways, and after its performance the other operation can be performed in any one “n2” ways, the number of ways of performing both the operations together is “n1 x n2”.

This theorem can be extended in case where more than two operations to be performed.

Ex. 1 : Two cards are drawn from a well shuffled pack of playing cards. Determine the probability that both are aces.

The total number of ways in which 2 cards can be drawn out of 52 is 52C2 = [(52X51) / (1X2)] = 26 x 51 = 1326

Since there are 4 aces in a pack of cards, so the number of favourable cases is 4C2 .

Hence, Required probability

= (4C2) / (52C2) = [{(4 x 3) / (1X 2)} / {(52 x 51) / (1 X 2)}] = [(4 x 3) / 2] x [2 /  (52 x 51)]

= 12 / (52 X 51) = 1/221

Ex. 2 : Four cards are drawn from a pack. Find the probability that two are spades and two are hearts.

Total number of possible selections (4 cards out of 52)= 52C4 .

Number of ways of selecting 2 spades cards  ( 2 cards out of 52) = 13C2 .

Number of ways of selecting 2 hearts cards  ( 2 cards out of 52) = 13C2 .

Total favourable cases = 13C2 x 13C2 .

So, the probability = [{13C2 x 13C2 } / 52C4 ] = [{(13X 12) / (2 x 1)} x {( 1 x 2 x 3 x 4)} / (52 x 51 x 50 x 49)}] = 468 / 20825

## Permutation and Combination – Problems

Statistical Permutation and Combination – Problems

A bag contains 5 white and 8 red balls. Two drawings of 3 balls are made such that (a) balls are replaced before the second trial and (b) the balls are not replaced before the second trial. Find the probability that the first drawing will give 3 white and the second 3 red balls in each case.

When the Balls are replaced

Possible cases: Selection of 3 balls out of 13 = 13C3 ways.

Favourable cases: Selection of 3 white balls out of 5 white balls in first draw = 5C3 .

Probability of 3 white in first trial =(5C3) / 13C3) = 5/ 143

Probability of 3 Red in second trial =(8C3) / 13C3) = 28/ 143

So, Probability of 3 white in first trial x Probability of 3 red in second trial =(5/ 143) x (28/ 143) = 140 / 20449

When the Balls are not replaced

Probability of drawing 3 white in first draw = 5/143

When the balls are not replaced, after the first draw, the total number of balls left = 2 white + 8 red = 10 balls.

In this case Probability of drawings 3 red balls in the second draw = (8C3) / (10C3) = 7/15

Required Compound Probability = (5/143) X (7/15) = 7/429

## Bernoulli’s Theorem of Probabilty

Bernoulli’s Theorem enumerates relative frequency of success in a sequence of trials

If the probability of occurring an event in one trial or experiment is known, then that the probability of happening of that event exactly r times out of n trials can be computed as

P(r) = nCr (p)r (q)n r, where P(r) = Probability of an event happening exactly r  times,

Where, n = No. of total Trials, r = Desired number of happening the event,  p = Probability of success or happening the event in one Trial, q = Probability of failure or not happening the event in one trial (1 – p)

Ex . Four coins are tossed simultaneously. What is the probability that there will be exactly two heads.

P(r) = nCr (p)r (q)n r, here n = 4, r = 2, p (Probability of head in throw of one coin) = 1/2

P(r) = 4C2 x (1/2) 2 x (1/2) 4-2 = [{(4x3x2) / (1×2)} x (1/4) x (1/4)}] = 3/8

## Statistical Expected Value of a Random Variable

Expected value or Mathematical or Expectation of a random variable may be defined as the sum of products of the different values taken by the random variable and the corresponding probabilities.

If a random variable x assumes n values x1, x2, x3 …., xn with corresponding probabilities p1, p2, p3 …., pn, then :

Expected value of x : m = E(x) = å PiXi , Expected value of x2 : E(x2) = åPixi2

Variance of x, denoted by s2 : V(x) =  E(X – m )2 = E(X2) – m 2

Properties of Expected Values

Expectation of a constant k is k, i.e. E(k) = k for any constant k.

Expectation of sum of two random variables is the sum of their expectations, i.e. E(x + y) = E(x) + E(y) for any two random variables x and y

Expectation off the product of a constant and a random variable is the product of the constant and the expectation of the random variable.  i.e. E(k x) = k.E(x) for any constant k

Expectation of the product of two random variables is the product of the two random variables, provided the two variables are independent., i.e. E(xy) = E(x) x E(y), Whenever x and y are independent.

## Expected Value : Problems

Statistical Expected Value – Problems

Ex. 1 : A die thrown at random. What is the expectations of number on it

Let X denote the number on the die. Then X is a random variable which can take any one of the values 1, 2, 3, …….. 6 each with equal probability 1/6 . Hence, E(X) = åx.p(x)

= {1 x (1/6)} + {2 x (1/6)} + {3 x (1/6)} + {4 x (1/6)} + {5 x (1/6)} + {6 x (1/6)}

= (1/6) x (1 + 2 + 3 + 4 + 5 + 6) = (1/6) x 21 = 21/6

Ex. 2. A random variable X has the following probability distribution: Find the value of K, expected value and variance of X.