Unit–1
Random variable and discrete probability distributions
Let A and B be two events of a sample space Sand let . Then conditional probability of the event A, given B, denoted byis defined by –
Theorem: If the events A and B defined on a sample space S of a random experiment are independent, then
Example1: A factory has two machines A and B making 60% and 40% respectively of the total production. Machine A produces 3% defective items, and B produces 5% defective items. Find the probability that a given defective part came from A.
SOLUTION: We consider the following events:
A: Selected item comes from A.
B: Selected item comes from B.
D: Selected item is defective.
We are looking for . We know:
Now,
So we need
Since, D is the union of the mutually exclusive events and (the entire sample space is the union of the mutually exclusive events A and B)
Example: Two fair dice are rolled, 1 red and 1 blue. The Sample Space is
S = {(1, 1),(1, 2), . . . ,(1, 6), . . . ,(6, 6)}.Total 36 outcomes, all equally likely (here (2, 3) denotes the outcome where the red die show 2 and the blue one shows 3).
(a) Consider the following events:
A: Red die shows 6.
B: Blue die shows 6.
Find , and .
Solution:
NOTE:so for this example. This is not surprising  we expect A to occur in of cases. In of these cases i.e. in of all cases, we expect B to also occur.
(b) Consider the following events:
C: Total Score is 10.
D: Red die shows an even number.
Find , and .
Solution:
NOTE:so,.
Why does multiplication not apply here as in part (a)?
ANSWER: Suppose C occurs: so the outcome is either (4, 6), (5, 5) or (6, 4). In two of these cases, namely (4, 6) and (6, 4), the event D also occurs. Thus
Although , the probability that D occurs given that C occurs is .
We write, and call the conditional probability of D given C.
NOTE: In the above example
Example: Three urns contain 6 red, 4 black; 4 red, 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red find the probability that it is drawn from the first urn.
Solution:
:The ball is drawn from urnI.
: The ball is drawn from urnII.
: The ball is drawn from urnIII.
R:The ball is red.
We have to find
Since the three urns are equally likely to be selected
Also,
From (i), we have
Example: A bag contains 12 pens of which 4 are defective. Three pens are picked at random from the bag one after the other.
Then find the probability that all three are nondefective.
Sol. Here the probability of the first which will be nondefective = 8/12
By the multiplication theorem of probability,
If we draw pens one after the other then the required probability will be
Example: The probability of A hits the target is 1 / 4 and the probability that B hits the target is 2/ 5. If both shoot the target then find the probability that at least one of them hits the target.
Sol.
Here it is given that
Now we have to find
Both two events are independent. So that
A probability space is a threetuple (S, F, P) in which the three components are
 Sample space: A nonempty set S called the sample space, which represents all possible outcomes.
 Event space: A collection F of subsets of S, called the event space.
 Probability function: A function P : FR, that assigns probabilities to the events in F.
DEFINITIONS:
1. Die: It is a small cube. Dots (number) are marked on its faces. Plural of the die is dice. On throwing a die, the outcome is the number of dots on its upper face.
2. Cards: A pack of cards consists of four suits i.e. Spades, Hearts, Diamonds and Clubs. Each suit consists of 13 cards, nine cards numbered 2, 3, 4, ..., 10, an Ace, a King, a Queen and a Jack or Knave. Colour of Spades and Clubs is black and that of Hearts and Diamonds is red.
Kings, Queens and Jacks are known as face cards.
3. Exhaustive Events or Sample Space: The set of all possible outcomes of a single performance of an experiment is exhaustive events or sample space. Each outcome is called a sample point.
In case of tossing a coin once, S = (H, T) is the sample space. Two outcomes  Head and Tail
 constitute an exhaustive event because no other outcome is possible.
4. Random Experiment: There are experiments, in which results may be altogether different, even though they are performed under identical conditions. They are known as random experiments. Tossing a coin or throwing a die is random experiment.
5. Trial and Event: Performing a random experiment is called a trial and outcome is termed as event. Tossing of a coin is a trial and the turning up of head or tail is an event.
6. Equally likely events: Two events are said to be ‘equally likely’, if one of them cannot be expected in preference to the other. For instance, if we draw a card from wellshuffled pack, we may get any card. Then the 52 different cases are equally likely.
7. Independent events: Two events may be independent, when the actual happening of one does not influence in any way the probability of the happening of the other.
8. Mutually Exclusive events: Two events are known as mutually exclusive, when the occurrence of one of them excludes the occurrence of the other. For example, on tossing of a coin, either we get head or tail, but not both.
9. Compound Event: When two or more events occur in composition with each other, the simultaneous occurrence is called a compound event. When a die is thrown, getting a 5 or 6 is a compound event.
10. Favourable Events: The events, which ensure the required happening, are said to be favourable events. For example, in throwing a die, to have the even numbers, 2, 4 and 6 are favourable cases.
11. Conditional Probability: The probability of happening an event A, such that event B has already happened, is called the conditional probability of happening of A on the condition that B has already happened. It is usually denoted by
12. Odds in favour of an event and odds against an event: If number of favourable ways = m, number of not favourable events = n
(i) Odds in favour of the event
(ii) Odds against the event
13. Classical Definition of Probability. If there are N equally likely, mutually, exclusive and exhaustive of events of an experiment and m of these are favourable, then the probability of the happening of the event is defined as
Consider Example1: In poker, a full house (3 cards of one rank and two of another, e.g. 3 fours and 2 queens) beats a flush (five cards of the same suit).
A player is more likely to be dealt a flush than a full house. We will be able to precisely quantify the meaning of “more likely” here.
Example2: A coin is tossed repeatedly.
Each toss has two possible outcomes:
Heads (H) or Tails (T)
Both equally likely. The outcome of each toss is unpredictable; so is the sequence of H and T.
However, as the number of tosses gets large, we expect that the number of H (heads) recorded will fluctuate around of the total number of tosses. We say the probability of aH is , abbreviated by . Of course also
Example3:
If 4 coins are tossed, what is the probability of getting 3 heads and 1 tail?
NOTES:
• In general, an event has associated to it a probability, which is a real number between 0 and 1.
• Events which are unlikely have low (close to 0) probability, and events which are likely have high (close to 1) probability.
• The probability of an event which is certain to occur is 1; the probability of an impossible event is 0.
Probability
Let S be a sample space, and each outcome of an experiment is equally likely and the number of outcomes is finite.
If A is an event connected with the experiment, then the probability of occurrence of event A is given by
Example: If a card is selected from the pack of playing cards then what will be the probability of the following cases?
1. A = {the card is a diamond}
2. B = {the card is a face card}
Sol.
As we know that
And \
Note
1. The probability of an event lies between 0 to 1, i.e.
2. If events A and B are mutually exclusive then
3. Let A be any event then
4. The sum of all probabilities in the sample space is 1.
5. The probability of an event which can not occur is 0.
Addition rule of probability
For any events A and B,
Example: In a college counselling, 30 students are choosing maths, 20 are choosing Computer science and 10 are selecting maths and computer science.
Suppose a student is selected randomly from 100 students then find the probability that student choose maths or computer science.
Sol.
Let A = {student is taking maths} and B = {student taking computer science}
So that the probabilities will be
Hence by the addition principle of probability
Independence:
Two events A, B ∈ are statistically independent iff
(Two disjoint events are not independent.)
Independence implies that
Knowing that outcome is in B does not change your perception of the outcome’s being in A.
Random Variable: It is a real valued function which assign a real number to each sample point in the sample space.
Random Variables
A random variable X is a function defined on the sample space 5 of an experiment S of an experiment. Its value are real numbers. For every number a the probability
With which X assumes a is defined. Similarly for any interval l the probability
With which X assumes any value in I is defined.
Example: 1.Tossing a fair coin thrice then
Sample Space(S) = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
2. Roll a dice
Sample Space(S) = {1,2,3,4,5,6}
Discrete Random Variable:
A random variable is said to be discrete if it has either a finite or a countable number of values
The number of students present each day in a class during an academic session is an example of discrete random variable as the number cannot take a fractional value.
Discrete Random Variables and Distribution
By definition a random variable X and its distribution are discrete if X assumes only finitely maany or at most contably many values called the possible values of X. With positive probabilities is zero for any interval J containing no possible values.
Clearly the discrete distribution of X is also determined by the probability functions f (x) of X, defined by
From this we get the values of the distribution function F (x) by taking sums.
Example: 1. No. Of head obtained when two coin are tossed.
2. No. Of defective items in a lot.
Probability mass function
Let X be a r.v. Which takes the values and let P[X = ] = p(. This function p(xi), i =1,2, … defined for the values assumed by X is called probability mass function of X satisfying p(xi) ≥ 0 and
Example: A random variable x has the following probability distribution
Then find
1. Value of c.
2. P[X≤3]
3. P[1 < X <4]
Sol.
We know that for the given probability distribution
So that
2.
3.
Independent random variables
Two discrete random variables X and Y are said to be independent only if
Note Two events are independent only if
Example: Two discrete random variables X and Y have
And
P[X = 1, Y = 1] = 5/9.
Check whether X and Y are independent or not?
Sol.
First we write the given distribution In tabular form
X/Y  0  1  P(x) 
0  2/9  1/9  3/9 
1  1/9  5/9  6/9 
P(y)  3/9  6/9  1 
Now
But
So that
Hence X and Y are not independent.
Let a random variable X has a probability distribution which assumes the values say with their associated probabilities then the mathematical expectation can be defined as
The expected value of a random variable X is written as E(X).
Expected value for a continuous random variable is
OR
The mean value (μ) of the probability distribution of a variate X is commonly known as its expectation current is denoted by E (X). If f(x) is the probability density function of the variate X, then
(discrete distribution)
(continuous distribution)
In general expectation of any function is given by
(discrete distribution)
(continuous distribution)
(2) Variance offer distribution is given by
(discrete distribution)
(continuous distribution)
Where is the standard deviation of the distribution.
(3) The rth moment about mean (denoted by is defined by
(discrete function)
(continuous function)
(4) Mean deviation from the mean is given by
(discrete distribution)
(continuous distribution)
Example. In a lottery, m tickets are drawn at a time out of a tickets numbered from 1 to n. Find the expected value of the sum of the numbers on the tickets drawn.
Solution. Let be the variables representing the numbers on the first, second,…nth ticket. The probability of drawing a ticket out of n ticket spelling in each case 1/n, we have
Therefore expected value of the sum of the numbers on the tickets drawn
Example. X is a continuous random variable with probability density function given by
Find k and mean value of X.
Solution. Since the total probability is unity.
Mean of X =
Example. The frequency distribution of a measurable characteristic varying between 0 and 2 is as under
Calculate two standard deviation and also the mean deviation about the mean.
Solution. Total frequency N =
(about the origin)=
(about the origin)=
Hence,
i.e., standard deviation
Mean derivation about the mean
Example: If a random variable X has the following probability distribution in tabular form then what will be the expected value of X.
X  0  1  2 
P(x)  1/4  1/2  1/4 
Sol.
We know that
So that
Example: Find the expectations of the number of an unbiased die when thrown.
Sol. Let X be a random variable which represents the number on a die when thrown.
X can take the values
1, 2, 3, 4, 5, 6
With
P[X = 1] = P[X = 2] = P[X = 3] = P[X = 4] = P[X = 5] = P[X = 6] = 1/6
The distribution table will be
X  1  2  3  4  5  6 
p(x)  1/6  1/6  1/6  1/6  1/6  1/6 
Hence the expectation of number on the die thrown is
So that
Independent random variables
In real life, we usually need to deal with more than one random variable. For example, if you study physical characteristics of people in a certain area you might pick a person at random and then look at his/ her weight, height etc. the weight of the randomly chosen person is one random variable while his/ her height is another one. Not only do we need to study is random variable separately but also we need to consider if there is dependence (i.e. correlation) between them. Is it true that a taller person is more likely to be heavier or not? the issues of dependence between several random variables will be studied in detail later on, but here we would like to talk about special scenario where two random variables are independent.
The concept of independent random variables is very similar to independent events. Remember to A and B are independent if we have P (A,B) = P (A) P (B) (remember comma means and, i.e. P(A,B) = P (A and B) = P (A B). Similarly we have the following definition for independent discrete random variables.
Definition
Consider two discrete random variables X and Y. We say that X and Y are independent if
In general if two random variables are independent then you can write
, For all sets A and B
Definition
Consider n discrete random variables We say that are independent if
Example.
I toss a coin twice and define X to be the number of heads one observe. Then I toss the coin two more times and define Y to be the number of heads that I observed this time. Find
Solution. since X and Y are the result of different independent coin tosses the two random variables X and Y are independent.
Example. A box A contains 2 white and 4 Black balls. Another box B contains 5 white and 7 black balls. A ball is transferred from the box A to the box B. Then A ball is drawn from the box B. Find the probability that it is white.
Solution. The probability of drawing a white ball from box B will depend on whether the transferred ball is black or white.
If black ball is transferred, it's probability is 4/6. There are now 5 white and 8 black balls in the box B.
Then the probability of drawing a white ball from box B is
Thus the probability of drawing a white ball from B, if the transferred ball is black
Similarly the probability of drawing a white ball from urn B if the transferred ball is white
Hens required probability
Example. A pair of dice is tossed twice. Find the probability of scoring 7 points
(a) Once
(b) At least once
(c) Twice
Solution. In a single toss of two dice the sum 7 can be obtained as (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), i.e. in 6 ways so that the probability of getting 7 = 6/36 = 1/6
Also the probability of not getting 7 = 11/6= 5/6
(a) The probability of getting seven in the first toss and not getting 7 in the second toss=1/6×5/6=5/36
Similarly the probability off not getting seven in the first toss and getting 7:00 in the second toss=5/6×1/6=5/36
Since these mutually exclusive events addition law of probability applies
Required probability
(b) The probability of not getting 7 in either toss
Therefore the probability of getting 7 at least once
(c) The probability of getting 7 twice
Example. Two cards are drawn in succession from a pack of 52 cards. Find the chance that the first is a king and the second a queen if the first card is
(i) Replaced
(ii) Got replaced
Solution. (i) the probability of drawing a king
If the card is replaced the back will again have 52 cards so that the probability of drawing a queen is 1/13.
The two events being independent the probability of drawing both cards in succession
(ii)the probability of drawing a king = 1/13
If the card is not replaced the pack will have 51 cards only so that the chance of drawing the queen is 4/51.
Hence the probability of drawing both cards
Example. Two cards are selected at random from 10 cards numbered 1to10 find the probability p that the sum is odd if
(i) The two cards are drawn together
(ii) the two cards are drawn one after the other without replacement
(iii) The two cards are drawn one after the other with replacement
Solution. (i) two cars out of 10 can be selected in ways. The sum is odd if one number is odd and other number is even. There being 5 odd numbers (1,3,5,7,9) and 5 even numbers (2,4,6,8,10) an what and an even number is chosen in 5×5=25 ways.
Thus,
(ii)two cards out of 10 can be selected one of the other without replacing in 10 × 10 = 100 ways
An odd number is selected in 5 × 5 =25 ways and an even number in 5 × 5 = 25 ways
Thus
(iii)two cards can be selected one after the other with replacement in 10 × 10 =100 ways
An odd number is selected in 5 × 5 =25 ways and an even number in 5×5=25 ways
Thus,
Moments
The r’th moment of a variable x about the mean is denoted by and defined as
The r’th moment of a variable x about any point ‘a’ will be
Relationship between moments about mean and moment about any point
When two variables are related in such a way that change in the value of one variable affects the value of the other variable, then these two variables are said to be correlated and there is correlation between two variables.
Example Height and weight of the persons of a group.
The correlation is said to be perfect correlation if two variables vary in such a way that their ratio is constant always.
Variance of a sum
One of the applications of covariance is finding the variance of a sum of several random variables. In particular, if Z = X + Y, then
Var (Z) =Cov (Z,Z)
More generally, for a, bR we conclude
Variance
Consider two random variables X and Y with the following PMFs
(3.3)
(3.4)
Note that EX =EY = 0. Although both random variables have the same mean value, their distribution is completely different. Y is always equal to its mean of 0, while X is IDA 100 or 100, quite far from its mean value. The variance is a measure of how spread out the distribution of a random variable is. Here the variance of Y is quite small since its distribution is concentrated value. Why the variance of X will be larger since its distribution is more spread out.
The variance of a random variable X with mean , is defined as
By definition the variance of X is the average value of Since ≥0, the variance is always larger than or equal to zero. A large value of the variance means that is often large, so X often X value far from its mean. This means that the distribution is very spread out. On the other hand a low variance means that the distribution is concentrated around its average.
Note that if we did not square the difference between X and its mean the result would be zero. That is
X is sometimes below its average and sometimes above its average. Thus is sometimes negative and sometimes positive but on average it is zero.
To compute , note that we need to find the expected value of , so we can use LOTUS. In particular we can write
For example, for X and Y defined in equations 3.3 and 3.4 we have
As we expect, X has a very large variance while Var (Y) = 0
Note that Var (X) has a different unit than X. For example, if X is measured in metres then Var(X) is in .to solve this issue we define another measure called the standard deviation usually shown as which is simply the square root of variance.
The standard deviation of a random variable X is defined as
The standard deviation of X has the same unit as X. For X and Y defined in equations 3.3 and 3.4 we have
Here is a useful formula for computing the variance.
Computational formula for the variance
To prove it note that
Note that for a given random variable X, is just a constant real number. Thus so we have
Equation 3.5 is equally easier to work with compared to . To use this equation we can find using LOTUS.
And then subtract to obtain the variance.
Example. I roll a fair die and let X be the resulting number. Find EX, Var(X), and
Solution. We have and for k = 1,2,…,6. Thus we have
Thus ,
Theorem
For random variable X and real number a and b
Proof. If
From equation 3.6, we conclude that, for standard deviation, . We mentioned that variance is NOT a linear operation. But there is a very important case, in which variance behave like a linear operation and that is when we look at sum of independent random variables,
Theorem
If are independent random variables and , then
Example. If Binomial (n, p) find Var (X).
Solution. We know that we can write a Binomial (n, p) random variable as the sum of n independent Bernoulli (p) random variable, i.e.
If Bernoulli (p) then its variance is
Problem. If , find Var (X).
Solution. We already know , thus Var (X). You can find directly using LOTUS, however, it is a little easier to find E [X (X1)] first. In particular using LOTUS we have
S
So we have . Thus, and we conclude
Karl Pearson’s coefficient of correlation
Here and
Note
1. Correlation coefficient always lies between 1 and +1.
2. Correlation coefficient is independent of change of origin and scale.
3. If the two variables are independent then correlation coefficient between them is zero.
Correlation coefficient  Type of correlation 
+1  Perfect positive correlation 
1  Perfect negative correlation 
0.25  Weak positive correlation 
0.75  Strong positive correlation 
0.25  Weak negative correlation 
0.75  Strong negative correlation 
0  No correlation 
Example: Find the correlation coefficient between Age and weight of the following data
Age  30  44  45  43  34  44 
Weight  56  55  60  64  62  63 
Sol.
x  y  ( ))  
30  56  10  100  4  16  40 
44  55  4  16  5  25  20 
45  60  5  25  0  0  0 
43  64  3  9  4  16  12 
34  62  6  36  2  4  12 
44  63  4  16  3  9  12 
Sum= 240 
360 
0 
202 
0 
70

32 
Karl Pearson’s coefficient of correlation
Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that the as age increases, the weight also increase.
Shortcut method to calculate correlation coefficient
Here,
Example: Find the correlation coefficient between the values X and Y of the dataset given below by using shortcut method
X  10  20  30  40  50 
Y  90  85  80  60  45 
Sol.
X  Y  
10  90  20  400  20  400  400 
20  85  10  100  15  225  150 
30  80  0  0  10  100  0 
40  60  10  100  10  100  100 
50  45  20  400  25  625  500 
Sum = 150 
360 
0 
1000 
10 
1450 
1150 
Shortcut method to calculate correlation coefficient
Spearman’s Rank Correlation
Solution: Let be the ranks of individuals corresponding to two characteristics.
Assuming nor two individuals are equal in either classification, each individual takes the values 1, 2, 3, and hence their arithmetic means are, each
Let , , , be the values of variable and , , those of
Then
Where and y are deviations from the mean.
Clearly, and
SPEARMAN’S RANK CORRELATION COEFFICIENT:
Where denotes rank coefficient of correlation and refers to the difference ofranks between paired items in two series.
Example: Compute Spearman’s rank correlation coefficient r for the following data:
Person  A  B  C  D  E  F  G  H  I  J 
Rank Statistics  9  10  6  5  7  2  4  8  1  3 
Rank in income  1  2  3  4  5  6  7  8  9  10 
Solution:
Person  Rank Statistics  Rank in income  d=  
A  9  1  8  64 
B  10  2  8  64 
C  6  3  3  9 
D  5  4  1  1 
E  7  5  2  4 
F  2  6  4  16 
G  4  7  3  9 
H  8  8  0  0 
I  1  9  8  64 
J  3  10  7  49 
Example: If X and Y are uncorrelated random variables, the of correlation between and
Solution:
Let and
Then
Now
Similarly
Now
Also
(As and are not correlated, we have )
Similarly
Markov and Chebyshev Inequalities
Let X be any positive continuous random variable, we can write
Thus, we conclude
We can prove the above inequality for discrete or mixed random variables similarly (using the generalized PDF), so we have the following result called Markov's inequality.
Markov Inequality
If X is any non negative random variable then
Example 1. Prove the union bound using Markov's inequality.
Solution. Similar to the discussion in the previous section, let be any events and X be the number events that occur. We saw that
Since X is a non negative random variable we can apply Markov's inequality. Choosing a = 1, we have
But note that )
Example 2. Let X ~ binomial (n, p). Using Markov's inequality, find an upper bound on P(X ≥n), where . Evaluate the ground for
Solution. Note that X is a non negative random variable and EX = np. Applying Markov's inequality we obtain
For ,we obtain
Chebyshev's Inequality
Let X be any random variable. If you define then Y is a nonnegative random variable so we can apply Markov''s inequality to Y. In particular for any positive real number b, we have
But note that
Thus we conclude that
This is Chebyshev’s inequality.
Chebyshev’s inequality
If X is any random variable then for any b>0 we have
Chebyshev’s inequality states that the difference between X and EX is somehow limited by Var (X). This is intuitively expected as variance shows on average how far we are from the mean.
Example 2. Let X ~ binomial (n, p). Using Markov's inequality, find an upper bound on P(X ≥n), where . Evaluate the ground for
Solution. One way to obtain a bound is to write
For p = ½ and , we obtain
Example. Let X be a random variables such that
Find a lower bound to its variance.
Solution.
The lower bound can be derived thanks to Chebyshev’s inequality
Thus, the lower bound is Var[X]≥2
Multinomial experiments consists of the following properties
1. The trials are independent.
2. Each trial has discrete number of possible outcomes.
3. The Experiments consists of n repeated trials.
Note Binomial distribution is a special case of multinomial distribution.
Multinomial distribution
A multinomial distribution is the probability distribution of the outcomes from a multinomial experiment.
The multinomial formula defines the probability of any outcome from a multinomial experiment.
Multinomial formula
Suppose there are n trials in a multinomial experiment and each trial can result in any of k possible outcomes . Now suppose each outcome can occur with probabilities .
Then the probability occurs times, occurs times and occurs times
Is
Example: If we have 10 balls in a bag, in which 2 are red, 3 are green and remaining are blue colour balls. Then we select 4 balls randomly form the bag with replacement, then what will be the probability of selecting 2 green balls and 2 blue balls.
Sol.
Here this experiment has 4 trials so n = 4.
The required probability will be
So that
Example: You are randomly drawing cards from an ordinary deck of cards. Every time you pick one you place it back in the deck. You do this 5 times. What is the probability of drawing 1 heart, 1 spade, 1club, and 2 diamonds?
Solution:
Example: A die weighed or loaded so that the number of spots X that appear on the up face when the die is rolled has pmf
If this loaded die is rolled 21 times. Find the probability of rolling one one, two twos, three threes, four fours, five fives, six sixes.
Solution:
Example: You are given a bag of marble. Inside the bag are 5 red marble, 4 white marble, 3 blue marble. Calculate the probability that with 6 trials you choose 3 marbles that are red, 1 marble that is white and 2 marble is blue. Replacing each marble after it is chosen.
Solution:
Poisson distribution
If it is a distribution related to the probabilities of events which are extremely rare but which have a large number of independent opportunities for occurrence. The number of persons born blind per year in a large city and the number of deaths by horse cake in an army corps are some of the phenomenon in which this law is followed.
This distribution can be derived as a limiting case of the binomial distribution of making a very large and p very small, keeping up fixed (=m, say).
The probability of r successes in a binomial distribution is
As (np=m), we have
So that the probability of 0,1,2…,r,… successors in a Poisson distribution are given by
The sum of these probabilities is unity as it should be.
(2) Constants of the Poisson distribution
These constants can easily be derived from the corresponding constants of the binomial distribution simply by making and noting that np =m
Standard deviation
Also
Skewness , Kurosis
Since is positive, Poisson distribution is positively skewed and since , it is Leptokurtic.
(3) Applications of poison distribution.
The distribution is applied to problems concerning:
(i) Arrival pattern of defective vehicles in a workshop, patients in a hospital or telephone calls.
(ii) Demand pattern for certain spare parts.
(iii) Number of fragments from a shell hitting a target.
(iv) Spatial distribution of bomb hits.
Example. If the probability of a bad reaction from a certain injection is 0.001, determine the chance that out of 2,000 individuals more than to get a bad reaction.
Solution. It follows a Poisson distribution as the probability of occurrence is very small
Mean m = np = (0.001)=2
Probability that more than two will get bad reaction
= 1 [probability that no one get a bad reaction + probability that one gets a bad reaction + probability that to get bad reaction]
Example. In a certain factory turning out razor blades there is a small chance of 0.002 of any need to be defective. The blades are supplied in packets of 10, use Poisson distribution to calculate the approximate number of packets containing no defective, one defective and to defective blades respectively in a consignment of 10000 packets.
Solution. We know that m = no = 10 × 0.002=0.02
Probability of no defective blade is
Therefore number of packets containing no defective blade is
Similarly the number of packets containing one defective blade
Find the number of packets containing to defective blades
Example. Fit a Poisson distribution to the set of observations:
x  0  1  2  3  4 
f  122  60  15  2  1 
Solution. Mean =
Therefore mean of Poisson distribution i.e. m =0.5
Hence the theoretical frequency for r successes is is
Therefore the theoretical frequencies are
x  0  1  2  3  4 
f  121  61  16  2  0 
A random experiment whose results are of only two types, for example, success S and failure F, is a Bernoulli test. The probability of success is taken as p, while the probability of failure is q = 1  p. Consider a random experiment of items in a sale, either sold or not sold. An item produced may be defective or nondefective. An egg is boiled or unboiled.
A random variable X will have the Bernoulli distribution with probability p if its probability distribution is
P(X = x) = px (1 – p)1−x, for x = 0, 1 and P(X = x) = 0 for other values of x.
Here, 0 is failure and 1 is success.
Conditions for Bernoulli tests
1. A finite number of tests.
2. Each trial must have exactly two results: success or failure.
3. The tests must be independent.
4. The probability of success or failure must be the same in each test.
Problem 1:
If the probability that a light bulb is defective is 0.8, what is the probability that the light bulb is not defective?
Solution:
Probability that the bulb is defective, p = 0.8
Probability that the bulb is not defective, q = 1  p = 1  0.8 = 0.2
Problem 2:
10 coins are tossed simultaneously where the probability of getting heads for each coin is 0.6. Find the probability of obtaining 4 heads.
Solution:
Probability of obtaining the head, p = 0.6
Probability of obtaining the head, q = 1  p = 1  0.6 = 0.4
Probability of obtaining 4 of 10 heads, P (X = 4) = C104 (0.6) 4 (0.4) 6P (X = 4) = C410 (0.6) 4 (0.4) 6 = 0.111476736
Problem 3:
In an exam, 10 multiplechoice questions are asked where only one in four answers is correct. Find the probability of getting 5 out of 10 correct questions on an answer sheet.
Solution:
Probability of obtaining a correct answer, p = 1414 = 0.25
Probability of obtaining a correct answer, q = 1  p = 1  0.25 = 0.75
Probability of obtaining 5 correct answers, P (X = 5) = C105 (0.25) 5 (0.75) 5C510 (0.25) 5 (0.75) 5 = 0.05839920044