Unit - 6

Probability

6.1 Random variable, discrete and continuous random variable, Probability density function

A random variable is a real-valued function whose domain is a set of possible outcomes of a random experiment and range is a sub-set of the set of real numbers and has the following properties:

i) Each particular value of the random variable can be assigned some probability

Ii) Uniting all the probabilities associated with all the different values of the random variable gives the value 1.

A random variable is denoted by X, Y, Z etc.

For example if a random experiment E consists of tossing a pair of dice, the sum X of the two numbers which turn up have the value 2,3,4,…12 depending on chance. Then X is a random variable 

Discrete random variable-

A random variable is said to be discrete if it has either a finite or a countable number of values. Countable number of values means the values which can be arranged in a sequence.

Note- if a random variable takes a finite set of values it is called discrete and if if a random variable takes an infinite number of uncountable values it is called continuous variable.

Discrete probability distributions-

Let X be a discrete variate which is the outcome of some experiments. If the probability that X takes the values of x is , then-

Where-

1.

2.

The set of values with their probabilities makes a discrete probability distribution of the discrete random variable X.

Probability distribution of a random variable X can be exhibited as follows-

X			and so on
P(x)

Example: Find the probability distribution of the number of heads when three coins are tossed simultaneously.

Sol.

Let be the number of heads in the toss of three coints

The sample space will be-

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Here variable X can take the values 0, 1, 2, 3 with the following probabilities-

P[X= 0] = P[TTT] = 1/8

P[X = 1] = P [HTT, THH, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P[HHH] = 1/8

Hence the probability distribution of X will be-

X
P(x)

Example: For the following probability distribution of a discrete random variable X,

Find-

1. The value of c.

2. P[1<x<4]

Sol,

We know that-

So that-

0 + c + c + 2c + 3c + c = 1

8c = 1

Then c = 1/8

Now, 2. P[1<x<4] = P[X = 2] + P[X = 3] = c + 2c = 3c = 3× 1/8 = 3/8

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-

P[X = 0, Y = 0] = 2/9 ,  P[X = 0, Y = 1] = 1/9,    P[X =1,  Y = 0] = 1/9

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-

Now-

But

So that-

Hence X and Y are not independent.

Continuous Random Variable:

A continuous random variable is a random variable where the data can take infinitely many values. For example, a random variable measuring the time taken for something to be done is continuous since there are an infinite number of possible times that can be taken.

Continuous random variable is called by a probability density function p (x), with given properties: p (x) ≥ 0 and the area between the x-axis  & the curve is 1: ... Standard deviation of a variable Random is defined by σ x = √Variance (x).

Continuous probability distribution (probability density function)-

When a random variable X takes every value in an interval, it gives to continuous distribution of X.

The probability distribution of a continuous variate x defined by a function f(x) such that the probabilities of the variate x falling in the interval

We express it as-

Here f(x) is called the probability density function.

Distribution function-

If F(x) = P(X≤x) = then F(x) is called as the distribution function of X.

It has the following properties-

1. F(-∞) = 0

2. F(∞) = 1

3.

Probability Density:

Probability density function (PDF) is a arithmetical appearance which gives a probability distribution for a discrete random variable as opposite to a continuous random variable. The difference among a discrete random variable is that we check an exact value of the variable. Like, the value for the variable, a stock worth, only goes two decimal points outside the decimal (Example 32.22), while a continuous variable have an countless number of values (Example 32.22564879…).

When the PDF is graphically characterized, the area under the curve will show the interval in which the variable will decline. The total area in this interval of the graph equals the probability of a discrete random variable happening. More exactly, since the absolute prospect of a continuous random variable taking on any exact value is zero owing to the endless set of possible values existing, the value of a PDF can be used to determine the likelihood of a random variable dropping within a exact range of values.

Probability density function-

Suppose f( x ) be a continuous function of x . Suppose the shaded region ABCD shown in the following figure represents the area bounded by y = f( x ), x –axis and the ordinates at the points x and x + δx , where δx is the length of the interval ( x , x + δx ).if δx is very-very small, then the curve AB will act as a line and hence the shaded region will be a rectangle whose area will be AD × DC this area = probability that X lies in the interval ( x, x +δx )

= P[ x≤ X ≤ x +δx ]

Hence,

Properties of Probability density function-

1. 

2.

Example: If a continuous random variable X has the following probability density function:

Then find-

1. P[0.2 < X < 0.5]

Sol.

Here f(x) is the probability density function, then-

Example. The probability density function of a variable X is

(i)                Find

(ii)             What will be e minimum value of k so that

Solution. (i) If X is a random variable then

(ii)Thus minimum value of k=1/30.

Example. A random variate X has the following probability function

x	0	1	2	3	4	5	6	7
P (x)	0	k	2k	2k	3k

(i)                Find the value of the k.

(ii)            

Solution. (i) If X is a random variable then

Example: Show that the following function can be defined as a density function and then find .

Sol.

Here

So that, the function can be defined as a density function.

Now.

Example:

Let X be a random variable with PDF given by

a, Find the constant c.

b. Find EX and Var (X).

c. Find P(X ).

Solution.

1. To find c, we can use

Thus we must have .

b.     To find EX we can write

In fact, we could have guessed EX = 0 because the PDF is symmetric around x = 0. To find Var (X) we have

c.      To find we can write

Example:

Let X be a continuous random variable with PDF given by

If , find the CDF of Y.

Solution. First we note that , we have

Thus,
 

Example:

Let X be a continuous random variable with PDF

Find .

Solution. We have

6.2 Binomial, Poisson and Normal distributions

BINOMIAL DISTRIBUTION

To find the probability of the happening of an event once, twice, thrice,…r times ….exactly in n trails.

Let the probability of the happening of an event A in one trial be p and its probability of not happening be 1 – p – q.

We assume that there are n trials and the happening of the event A is r times and its not happening is n – r times.

This may be shown as follows

AA……A                                 

r times                                       n – r times                          (1)

A indicates its happening its failure and P (A) =p and P (

We see that (1) has the probability

Pp…p qq….q=

r times              n-r times                                                       (2)

Clearly (1) is merely one order of arranging r A’S.

The probability of (1) =Number of different arrangements of r A’s and (n-r)’s

The number of different arrangements of r A’s and (n-r)’s

Probability of the happening of an event r times =

If r = 0, probability of happening of an event 0 times

If r = 1,probability of happening of an event 1 times

If r = 2,probability of happening of an event 2 times

If r = 3,probability of happening of an event 3 times and so on.

These terms are clearly the successive terms in the expansion of

Hence it is called Binomial Distribution.

Example. If on an average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.

Solution:

Out of 10 ships one ship is wrecked.

I.e. nine ships out of 10 ships are safe, P (safety) =

P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)

Example. The overall percentage of failures in a certain examination is 20. If 6 candidates appear in the examination what is the probability that at least five pass the examination?

Solution:

Probability of failures = 20%

Probability of (P) =

Probability of at least 5 pass = P(5 or 6)

Example:

The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now 60, at least seven will live to be 70?

Solution:

The probability that a man aged 60 will live to be 70

Number of men= n = 10

Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)

= P (7)+ P(8)+ P(9) + P(10) =

Example

Assuming that 20% of the population of a city are literate so that the chance of an individual being literate is and assuming that hundred investigators each take 10 individuals to see whether they are illiterate, how many investigators would you expect to report 3 or less were literate.

Solution

Required number of investigators = 0.879126118× 100 =87.9126118

= 88 approximate

Mean or binomial distribution

Successors r	Frequency f	Rf
0		0
1
2		n(n-1)
3
…..	……	….
n

Since,

STANDARD DEVIATION OF BINOMIAL DISTRIBUTION

Successors r	Frequency f
0		0
1
2		2n(n-1)
3
…..	……	….
n

We know that           (1)

r is the deviation of items (successes) from 0.

Putting these values in (1) we have

Hence for the binomial distribution, Mean

Example. A die is tossed thrice. A success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.

Solution.

RECURRENCE RELATION FOR THE BINOMIAL DISTRIBUTION

By Binomial Distribution

On dividing (2) by (1) , we get

Poisson Distribution:

Poisson distribution is a particular limiting form of the Binomial distribution when p (or q) is very small and n is large enough.

Poisson distribution is

Where m is the mean of the distribution.

Proof. In Binomial Distribution

Taking limits when n tends to infinity

MEAN OF POISSON DISTRIBUTION

Success r	Frequency f	f.r
0		0
1
2
3
…	…	…
r
…	…	…

STANDARD DEVIATION OF POISSON DISTRIBUTION

Successive r	Frequency f	Product rf	Product
0		0	0
1
2
3
…….	……..	……..	……..
r
……..	…….	……..	…….

Hence mean and variance of a Poisson distribution are equal to m. Similarly we can obtain,

MEAN DEVIATION

Show that in a Poisson distribution with unit mean, and the mean deviation about the mean is 2/e times the standard deviation.

Solution

  But mean = 1 i.e. m =1 and S.D. =

r	P (r)	|r-1|	P(r)|r-1|
0		1
1		0	0
2		1
3		2
4		3
…..	…..	…..	…..
r		r-1

Mean Deviation =

MOMENT GENERATING FUNCTION OF POISSON DISTRIBUTION

Solution

Let be the moment generating function then

CUMULANTS

The cumulant generating function is given by

Now cumulant =coefficient of in K (t) = m

i.e. , where r = 1,2,3,…

Mean =

RECURRENCE FORMULA FOR POISSON DISTRIBUTION

SOLUTION.

By Poisson distribution

On dividing (2) by (1) we get

Example. Assume that the probability of an individual coal miner being killed in a mine accident during a year is . Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.

Solution

Example. Suppose 3% of bolts made by a machine are defective, the defects occuring at random during production. If bolts are packaged 50 per box, find

(a)  Exact probability and

(b)  Poisson approximation to it, that a given box will contain 5 defectives.

Solution

(a)  Hence the probability for 5 defectives bolts in a lot of 50.

(b) To get Poisson approximation m = np =

Required Poisson approximation=

Example. In a certain factory producing cycle tyres, there is a smallchance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective, one defective and two defective tyres, respectively, in a consignment of 10,000 lots.

Solution.

S.No.	Probability of defective	Number of lots containing defective
1.
2.
3.

Normal Distribution-

The concept of normal distribution was given by English mathematician Abraham De Moivre in 1733 but the concrete theory was given by Karl Gauss that is why sometime normal distribution is called Gaussian distribution.

Normal distribution is a continuous distribution. It is a limiting case of binomial distribution.

The probability density function of a normal distribution is given by-

Here

Where

Note-

1. If a random variable X follows normal distribution with mean and variance then we can write it as- X

2. If X , then is called standard normal variate with mean 0 and standard deviation 1.

3. The probability density function of standard normal variate Z is given as-

Where

Graph of a normal probability function-

The curve look like bell-shaped curve. The top of the bell is exactly above the mean.

If the value of standard deviation is large then curve tends to flatten out and for small standard deviation it has sharp peak.

This is one of the most important probability distributions in statistical analysis.

Example:

1. If X then find the probability density function of X.

2. If X then find the probability density function of X.

Sol.

1. We are given X

Here

We know that-

Then the p.d.f.  will be-

2. . We are given X

Here

We know that-

Then the p.d.f.  will be-

Mean, median and mode of the normal distribution-

Let ‘a’ is the median, then it divides the total area into two parts-

Where-

Let a>mean, then-

Thus-

So that mean = median.

Note- mean deviation about mean is =

Mode-

The mode of the normal distribution is and modal ordinate is given by-

Hence the mean, median and mode are equal in normal distribution.

Area property of a normal distribution (Area under the normal curve)-

Let X follows the normal distribution with mean and variance

We form a normal curve by taking

Note- Total area under the curve is always 1.

Example: If a random variable X is normally distributed with mean 80 and standard deviation 5, then find-

1. P[X > 95]

2. P[X < 72]

3. P [85 < X <97]

[Note- use the table- area under the normal curve]

Sol.

The standard normal variate is –

Now-

1. X = 95,

So that-

2. X = 72,

So that-

3. X = 85,

X = 97,

So that-

Example: In a company the mean weight of 1000 employees is 60kg and standard deviation is 16kg.

Find the number of employees having their weights-

1. Less than 55kg.

2. More than 70kg.

3. Between 45kg and 65kg.

Sol. Suppose X be a normal variate = the weight of employees.

Here mean 60kg and S.D. = 16kg

X

Then we know that-

We get from the data,

Now-

1. For X = 55,

So that-

P[X < 55] = P[Z < - 0.31] = P[Z > 0.31]

= 0.5 – P [ 0 < Z < 0.31]

= 0.5 – 0.1217

2. For X = 70,

So that-

P[X > 70] = P[Z > 0.63]

= 0.5 – P[0 < Z < 0.63]

= 0.5 – 0.2357 = 0.2643

3. For X = 45,

For X = 65,

Hence the number of employees having weights between 45kg and 65kg-

Example: The mean inside diameter of a sample of 200 washers produced by a machine is 0.0502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

Sol.

Here-

And

Area for non-defective washers = area between z = -1.2 to +1.2

= 2 area between z = 0 and z = 1.2

= 2 × 0.3849 = 0.7698 = 76.98%

Then percent of defective washers = 100 – 76.98 = 23.02 %

Example: The life of electric bulbs is normally distributed with mean 8 months and standard deviation 2 months.

If 5000 electric bulbs are issued how many bulbs should be expected to need replacement after 12 months?

[Given that P (z ≥ 2) = 0. 0228]

Sol.

Here mean (μ) = 8 and standard deviation = 2

Number of bulbs = 5000

Total months (X) = 12

We know that-

Area (z ≥ 2) = 0.0228

Number of electric bulbs whose life is more than 12 months ( Z> 12)

= 5000 × 0.0228 = 114

Therefore replacement after 12 months = 5000 – 114 = 4886 electric bulbs.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 6

Probability

6.1 Random variable, discrete and continuous random variable, Probability density function

A random variable is a real-valued function whose domain is a set of possible outcomes of a random experiment and range is a sub-set of the set of real numbers and has the following properties:

i) Each particular value of the random variable can be assigned some probability

Ii) Uniting all the probabilities associated with all the different values of the random variable gives the value 1.

A random variable is denoted by X, Y, Z etc.

For example if a random experiment E consists of tossing a pair of dice, the sum X of the two numbers which turn up have the value 2,3,4,…12 depending on chance. Then X is a random variable 

Discrete random variable-

A random variable is said to be discrete if it has either a finite or a countable number of values. Countable number of values means the values which can be arranged in a sequence.

Note- if a random variable takes a finite set of values it is called discrete and if if a random variable takes an infinite number of uncountable values it is called continuous variable.

Discrete probability distributions-

Let X be a discrete variate which is the outcome of some experiments. If the probability that X takes the values of x is , then-

Where-

1.

2.

The set of values with their probabilities makes a discrete probability distribution of the discrete random variable X.

Probability distribution of a random variable X can be exhibited as follows-

X			and so on
P(x)

Example: Find the probability distribution of the number of heads when three coins are tossed simultaneously.

Sol.

Let be the number of heads in the toss of three coints

The sample space will be-

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Here variable X can take the values 0, 1, 2, 3 with the following probabilities-

P[X= 0] = P[TTT] = 1/8

P[X = 1] = P [HTT, THH, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P[HHH] = 1/8

Hence the probability distribution of X will be-

X
P(x)

Example: For the following probability distribution of a discrete random variable X,

Find-

1. The value of c.

2. P[1<x<4]

Sol,

We know that-

So that-

0 + c + c + 2c + 3c + c = 1

8c = 1

Then c = 1/8

Now, 2. P[1<x<4] = P[X = 2] + P[X = 3] = c + 2c = 3c = 3× 1/8 = 3/8

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-

P[X = 0, Y = 0] = 2/9 ,  P[X = 0, Y = 1] = 1/9,    P[X =1,  Y = 0] = 1/9

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-

Now-

But

So that-

Hence X and Y are not independent.

Continuous Random Variable:

A continuous random variable is a random variable where the data can take infinitely many values. For example, a random variable measuring the time taken for something to be done is continuous since there are an infinite number of possible times that can be taken.

Continuous random variable is called by a probability density function p (x), with given properties: p (x) ≥ 0 and the area between the x-axis  & the curve is 1: ... Standard deviation of a variable Random is defined by σ x = √Variance (x).

Continuous probability distribution (probability density function)-

When a random variable X takes every value in an interval, it gives to continuous distribution of X.

The probability distribution of a continuous variate x defined by a function f(x) such that the probabilities of the variate x falling in the interval

We express it as-

Here f(x) is called the probability density function.

Distribution function-

If F(x) = P(X≤x) = then F(x) is called as the distribution function of X.

It has the following properties-

1. F(-∞) = 0

2. F(∞) = 1

3.

Probability Density:

Probability density function (PDF) is a arithmetical appearance which gives a probability distribution for a discrete random variable as opposite to a continuous random variable. The difference among a discrete random variable is that we check an exact value of the variable. Like, the value for the variable, a stock worth, only goes two decimal points outside the decimal (Example 32.22), while a continuous variable have an countless number of values (Example 32.22564879…).

When the PDF is graphically characterized, the area under the curve will show the interval in which the variable will decline. The total area in this interval of the graph equals the probability of a discrete random variable happening. More exactly, since the absolute prospect of a continuous random variable taking on any exact value is zero owing to the endless set of possible values existing, the value of a PDF can be used to determine the likelihood of a random variable dropping within a exact range of values.

Probability density function-

Suppose f( x ) be a continuous function of x . Suppose the shaded region ABCD shown in the following figure represents the area bounded by y = f( x ), x –axis and the ordinates at the points x and x + δx , where δx is the length of the interval ( x , x + δx ).if δx is very-very small, then the curve AB will act as a line and hence the shaded region will be a rectangle whose area will be AD × DC this area = probability that X lies in the interval ( x, x +δx )

= P[ x≤ X ≤ x +δx ]

Hence,

Properties of Probability density function-

1. 

2.

Example: If a continuous random variable X has the following probability density function:

Then find-

1. P[0.2 < X < 0.5]

Sol.

Here f(x) is the probability density function, then-

Example. The probability density function of a variable X is

(i)                Find

(ii)             What will be e minimum value of k so that

Solution. (i) If X is a random variable then

(ii)Thus minimum value of k=1/30.

Example. A random variate X has the following probability function

x	0	1	2	3	4	5	6	7
P (x)	0	k	2k	2k	3k

(i)                Find the value of the k.

(ii)            

Solution. (i) If X is a random variable then

Example: Show that the following function can be defined as a density function and then find .

Sol.

Here

So that, the function can be defined as a density function.

Now.

Example:

Let X be a random variable with PDF given by

a, Find the constant c.

b. Find EX and Var (X).

c. Find P(X ).

Solution.

1. To find c, we can use

Thus we must have .

b.     To find EX we can write

In fact, we could have guessed EX = 0 because the PDF is symmetric around x = 0. To find Var (X) we have

c.      To find we can write

Example:

Let X be a continuous random variable with PDF given by

If , find the CDF of Y.

Solution. First we note that , we have

Thus,
 

Example:

Let X be a continuous random variable with PDF

Find .

Solution. We have

6.2 Binomial, Poisson and Normal distributions

BINOMIAL DISTRIBUTION

To find the probability of the happening of an event once, twice, thrice,…r times ….exactly in n trails.

Let the probability of the happening of an event A in one trial be p and its probability of not happening be 1 – p – q.

We assume that there are n trials and the happening of the event A is r times and its not happening is n – r times.

This may be shown as follows

AA……A                                 

r times                                       n – r times                          (1)

A indicates its happening its failure and P (A) =p and P (

We see that (1) has the probability

Pp…p qq….q=

r times              n-r times                                                       (2)

Clearly (1) is merely one order of arranging r A’S.

The probability of (1) =Number of different arrangements of r A’s and (n-r)’s

The number of different arrangements of r A’s and (n-r)’s

Probability of the happening of an event r times =

If r = 0, probability of happening of an event 0 times

If r = 1,probability of happening of an event 1 times

If r = 2,probability of happening of an event 2 times

If r = 3,probability of happening of an event 3 times and so on.

These terms are clearly the successive terms in the expansion of

Hence it is called Binomial Distribution.

Example. If on an average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.

Solution:

Out of 10 ships one ship is wrecked.

I.e. nine ships out of 10 ships are safe, P (safety) =

P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)

Example. The overall percentage of failures in a certain examination is 20. If 6 candidates appear in the examination what is the probability that at least five pass the examination?

Solution:

Probability of failures = 20%

Probability of (P) =

Probability of at least 5 pass = P(5 or 6)

Example:

The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now 60, at least seven will live to be 70?

Solution:

The probability that a man aged 60 will live to be 70

Number of men= n = 10

Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)

= P (7)+ P(8)+ P(9) + P(10) =

Example

Assuming that 20% of the population of a city are literate so that the chance of an individual being literate is and assuming that hundred investigators each take 10 individuals to see whether they are illiterate, how many investigators would you expect to report 3 or less were literate.

Solution

Required number of investigators = 0.879126118× 100 =87.9126118

= 88 approximate

Mean or binomial distribution

Successors r	Frequency f	Rf
0		0
1
2		n(n-1)
3
…..	……	….
n

Since,

STANDARD DEVIATION OF BINOMIAL DISTRIBUTION

Successors r	Frequency f
0		0
1
2		2n(n-1)
3
…..	……	….
n

We know that           (1)

r is the deviation of items (successes) from 0.

Putting these values in (1) we have

Hence for the binomial distribution, Mean

Example. A die is tossed thrice. A success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.

Solution.

RECURRENCE RELATION FOR THE BINOMIAL DISTRIBUTION

By Binomial Distribution

On dividing (2) by (1) , we get

Poisson Distribution:

Poisson distribution is a particular limiting form of the Binomial distribution when p (or q) is very small and n is large enough.

Poisson distribution is

Where m is the mean of the distribution.

Proof. In Binomial Distribution

Taking limits when n tends to infinity

MEAN OF POISSON DISTRIBUTION

Success r	Frequency f	f.r
0		0
1
2
3
…	…	…
r
…	…	…

STANDARD DEVIATION OF POISSON DISTRIBUTION

Successive r	Frequency f	Product rf	Product
0		0	0
1
2
3
…….	……..	……..	……..
r
……..	…….	……..	…….

Hence mean and variance of a Poisson distribution are equal to m. Similarly we can obtain,

MEAN DEVIATION

Show that in a Poisson distribution with unit mean, and the mean deviation about the mean is 2/e times the standard deviation.

Solution

  But mean = 1 i.e. m =1 and S.D. =

r	P (r)	|r-1|	P(r)|r-1|
0		1
1		0	0
2		1
3		2
4		3
…..	…..	…..	…..
r		r-1

Mean Deviation =

MOMENT GENERATING FUNCTION OF POISSON DISTRIBUTION

Solution

Let be the moment generating function then

CUMULANTS

The cumulant generating function is given by

Now cumulant =coefficient of in K (t) = m

i.e. , where r = 1,2,3,…

Mean =

RECURRENCE FORMULA FOR POISSON DISTRIBUTION

SOLUTION.

By Poisson distribution

On dividing (2) by (1) we get

Example. Assume that the probability of an individual coal miner being killed in a mine accident during a year is . Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.

Solution

Example. Suppose 3% of bolts made by a machine are defective, the defects occuring at random during production. If bolts are packaged 50 per box, find

(a)  Exact probability and

(b)  Poisson approximation to it, that a given box will contain 5 defectives.

Solution

(a)  Hence the probability for 5 defectives bolts in a lot of 50.

(b) To get Poisson approximation m = np =

Required Poisson approximation=

Example. In a certain factory producing cycle tyres, there is a smallchance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective, one defective and two defective tyres, respectively, in a consignment of 10,000 lots.

Solution.

S.No.	Probability of defective	Number of lots containing defective
1.
2.
3.

Normal Distribution-

The concept of normal distribution was given by English mathematician Abraham De Moivre in 1733 but the concrete theory was given by Karl Gauss that is why sometime normal distribution is called Gaussian distribution.

Normal distribution is a continuous distribution. It is a limiting case of binomial distribution.

The probability density function of a normal distribution is given by-

Here

Where

Note-

1. If a random variable X follows normal distribution with mean and variance then we can write it as- X

2. If X , then is called standard normal variate with mean 0 and standard deviation 1.

3. The probability density function of standard normal variate Z is given as-

Where

Graph of a normal probability function-

The curve look like bell-shaped curve. The top of the bell is exactly above the mean.

If the value of standard deviation is large then curve tends to flatten out and for small standard deviation it has sharp peak.

This is one of the most important probability distributions in statistical analysis.

Example:

1. If X then find the probability density function of X.

2. If X then find the probability density function of X.

Sol.

1. We are given X

Here

We know that-

Then the p.d.f.  will be-

2. . We are given X

Here

We know that-

Then the p.d.f.  will be-

Mean, median and mode of the normal distribution-

Let ‘a’ is the median, then it divides the total area into two parts-

Where-

Let a>mean, then-

Thus-

So that mean = median.

Note- mean deviation about mean is =

Mode-

The mode of the normal distribution is and modal ordinate is given by-

Hence the mean, median and mode are equal in normal distribution.

Area property of a normal distribution (Area under the normal curve)-

Let X follows the normal distribution with mean and variance

We form a normal curve by taking

Note- Total area under the curve is always 1.

Example: If a random variable X is normally distributed with mean 80 and standard deviation 5, then find-

1. P[X > 95]

2. P[X < 72]

3. P [85 < X <97]

[Note- use the table- area under the normal curve]

Sol.

The standard normal variate is –

Now-

1. X = 95,

So that-

2. X = 72,

So that-

3. X = 85,

X = 97,

So that-

Example: In a company the mean weight of 1000 employees is 60kg and standard deviation is 16kg.

Find the number of employees having their weights-

1. Less than 55kg.

2. More than 70kg.

3. Between 45kg and 65kg.

Sol. Suppose X be a normal variate = the weight of employees.

Here mean 60kg and S.D. = 16kg

X

Then we know that-

We get from the data,

Now-

1. For X = 55,

So that-

P[X < 55] = P[Z < - 0.31] = P[Z > 0.31]

= 0.5 – P [ 0 < Z < 0.31]

= 0.5 – 0.1217

2. For X = 70,

So that-

P[X > 70] = P[Z > 0.63]

= 0.5 – P[0 < Z < 0.63]

= 0.5 – 0.2357 = 0.2643

3. For X = 45,

For X = 65,

Hence the number of employees having weights between 45kg and 65kg-

Example: The mean inside diameter of a sample of 200 washers produced by a machine is 0.0502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

Sol.

Here-

And

Area for non-defective washers = area between z = -1.2 to +1.2

= 2 area between z = 0 and z = 1.2

= 2 × 0.3849 = 0.7698 = 76.98%

Then percent of defective washers = 100 – 76.98 = 23.02 %

Example: The life of electric bulbs is normally distributed with mean 8 months and standard deviation 2 months.

If 5000 electric bulbs are issued how many bulbs should be expected to need replacement after 12 months?

[Given that P (z ≥ 2) = 0. 0228]

Sol.

Here mean (μ) = 8 and standard deviation = 2

Number of bulbs = 5000

Total months (X) = 12

We know that-

Area (z ≥ 2) = 0.0228

Number of electric bulbs whose life is more than 12 months ( Z> 12)

= 5000 × 0.0228 = 114

Therefore replacement after 12 months = 5000 – 114 = 4886 electric bulbs.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 6

Probability

6.1 Random variable, discrete and continuous random variable, Probability density function

A random variable is a real-valued function whose domain is a set of possible outcomes of a random experiment and range is a sub-set of the set of real numbers and has the following properties:

i) Each particular value of the random variable can be assigned some probability

Ii) Uniting all the probabilities associated with all the different values of the random variable gives the value 1.

A random variable is denoted by X, Y, Z etc.

For example if a random experiment E consists of tossing a pair of dice, the sum X of the two numbers which turn up have the value 2,3,4,…12 depending on chance. Then X is a random variable 

Discrete random variable-

A random variable is said to be discrete if it has either a finite or a countable number of values. Countable number of values means the values which can be arranged in a sequence.

Note- if a random variable takes a finite set of values it is called discrete and if if a random variable takes an infinite number of uncountable values it is called continuous variable.

Discrete probability distributions-

Let X be a discrete variate which is the outcome of some experiments. If the probability that X takes the values of x is , then-

Where-

1.

2.

The set of values with their probabilities makes a discrete probability distribution of the discrete random variable X.

Probability distribution of a random variable X can be exhibited as follows-

X			and so on
P(x)

Example: Find the probability distribution of the number of heads when three coins are tossed simultaneously.

Sol.

Let be the number of heads in the toss of three coints

The sample space will be-

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Here variable X can take the values 0, 1, 2, 3 with the following probabilities-

P[X= 0] = P[TTT] = 1/8

P[X = 1] = P [HTT, THH, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P[HHH] = 1/8

Hence the probability distribution of X will be-

X
P(x)

Example: For the following probability distribution of a discrete random variable X,

Find-

1. The value of c.

2. P[1<x<4]

Sol,

We know that-

So that-

0 + c + c + 2c + 3c + c = 1

8c = 1

Then c = 1/8

Now, 2. P[1<x<4] = P[X = 2] + P[X = 3] = c + 2c = 3c = 3× 1/8 = 3/8

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-

P[X = 0, Y = 0] = 2/9 ,  P[X = 0, Y = 1] = 1/9,    P[X =1,  Y = 0] = 1/9

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-

Now-

But

So that-

Hence X and Y are not independent.

Continuous Random Variable:

A continuous random variable is a random variable where the data can take infinitely many values. For example, a random variable measuring the time taken for something to be done is continuous since there are an infinite number of possible times that can be taken.

Continuous random variable is called by a probability density function p (x), with given properties: p (x) ≥ 0 and the area between the x-axis  & the curve is 1: ... Standard deviation of a variable Random is defined by σ x = √Variance (x).

Continuous probability distribution (probability density function)-

When a random variable X takes every value in an interval, it gives to continuous distribution of X.

The probability distribution of a continuous variate x defined by a function f(x) such that the probabilities of the variate x falling in the interval

We express it as-

Here f(x) is called the probability density function.

Distribution function-

If F(x) = P(X≤x) = then F(x) is called as the distribution function of X.

It has the following properties-

1. F(-∞) = 0

2. F(∞) = 1

3.

Probability Density:

Probability density function (PDF) is a arithmetical appearance which gives a probability distribution for a discrete random variable as opposite to a continuous random variable. The difference among a discrete random variable is that we check an exact value of the variable. Like, the value for the variable, a stock worth, only goes two decimal points outside the decimal (Example 32.22), while a continuous variable have an countless number of values (Example 32.22564879…).

When the PDF is graphically characterized, the area under the curve will show the interval in which the variable will decline. The total area in this interval of the graph equals the probability of a discrete random variable happening. More exactly, since the absolute prospect of a continuous random variable taking on any exact value is zero owing to the endless set of possible values existing, the value of a PDF can be used to determine the likelihood of a random variable dropping within a exact range of values.

Probability density function-

Suppose f( x ) be a continuous function of x . Suppose the shaded region ABCD shown in the following figure represents the area bounded by y = f( x ), x –axis and the ordinates at the points x and x + δx , where δx is the length of the interval ( x , x + δx ).if δx is very-very small, then the curve AB will act as a line and hence the shaded region will be a rectangle whose area will be AD × DC this area = probability that X lies in the interval ( x, x +δx )

= P[ x≤ X ≤ x +δx ]

Hence,

Properties of Probability density function-

1. 

2.

Example: If a continuous random variable X has the following probability density function:

Then find-

1. P[0.2 < X < 0.5]

Sol.

Here f(x) is the probability density function, then-

Example. The probability density function of a variable X is

(i)                Find

(ii)             What will be e minimum value of k so that

Solution. (i) If X is a random variable then

(ii)Thus minimum value of k=1/30.

Example. A random variate X has the following probability function

x	0	1	2	3	4	5	6	7
P (x)	0	k	2k	2k	3k

(i)                Find the value of the k.

(ii)            

Solution. (i) If X is a random variable then

Example: Show that the following function can be defined as a density function and then find .

Sol.

Here

So that, the function can be defined as a density function.

Now.

Example:

Let X be a random variable with PDF given by

a, Find the constant c.

b. Find EX and Var (X).

c. Find P(X ).

Solution.

1. To find c, we can use

Thus we must have .

b.     To find EX we can write

In fact, we could have guessed EX = 0 because the PDF is symmetric around x = 0. To find Var (X) we have

c.      To find we can write

Example:

Let X be a continuous random variable with PDF given by

If , find the CDF of Y.

Solution. First we note that , we have

Thus,
 

Example:

Let X be a continuous random variable with PDF

Find .

Solution. We have

6.2 Binomial, Poisson and Normal distributions

BINOMIAL DISTRIBUTION

To find the probability of the happening of an event once, twice, thrice,…r times ….exactly in n trails.

Let the probability of the happening of an event A in one trial be p and its probability of not happening be 1 – p – q.

We assume that there are n trials and the happening of the event A is r times and its not happening is n – r times.

This may be shown as follows

AA……A                                 

r times                                       n – r times                          (1)

A indicates its happening its failure and P (A) =p and P (

We see that (1) has the probability

Pp…p qq….q=

r times              n-r times                                                       (2)

Clearly (1) is merely one order of arranging r A’S.

The probability of (1) =Number of different arrangements of r A’s and (n-r)’s

The number of different arrangements of r A’s and (n-r)’s

Probability of the happening of an event r times =

If r = 0, probability of happening of an event 0 times

If r = 1,probability of happening of an event 1 times

If r = 2,probability of happening of an event 2 times

If r = 3,probability of happening of an event 3 times and so on.

These terms are clearly the successive terms in the expansion of

Hence it is called Binomial Distribution.

Example. If on an average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.

Solution:

Out of 10 ships one ship is wrecked.

I.e. nine ships out of 10 ships are safe, P (safety) =

P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)

Example. The overall percentage of failures in a certain examination is 20. If 6 candidates appear in the examination what is the probability that at least five pass the examination?

Solution:

Probability of failures = 20%

Probability of (P) =

Probability of at least 5 pass = P(5 or 6)

Example:

The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now 60, at least seven will live to be 70?

Solution:

The probability that a man aged 60 will live to be 70

Number of men= n = 10

Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)

= P (7)+ P(8)+ P(9) + P(10) =

Example

Assuming that 20% of the population of a city are literate so that the chance of an individual being literate is and assuming that hundred investigators each take 10 individuals to see whether they are illiterate, how many investigators would you expect to report 3 or less were literate.

Solution

Required number of investigators = 0.879126118× 100 =87.9126118

= 88 approximate

Mean or binomial distribution

Successors r	Frequency f	Rf
0		0
1
2		n(n-1)
3
…..	……	….
n

Since,

STANDARD DEVIATION OF BINOMIAL DISTRIBUTION

Successors r	Frequency f
0		0
1
2		2n(n-1)
3
…..	……	….
n

We know that           (1)

r is the deviation of items (successes) from 0.

Putting these values in (1) we have

Hence for the binomial distribution, Mean

Example. A die is tossed thrice. A success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.

Solution.

RECURRENCE RELATION FOR THE BINOMIAL DISTRIBUTION

By Binomial Distribution

On dividing (2) by (1) , we get

Poisson Distribution:

Poisson distribution is a particular limiting form of the Binomial distribution when p (or q) is very small and n is large enough.

Poisson distribution is

Where m is the mean of the distribution.

Proof. In Binomial Distribution

Taking limits when n tends to infinity

MEAN OF POISSON DISTRIBUTION

Success r	Frequency f	f.r
0		0
1
2
3
…	…	…
r
…	…	…

STANDARD DEVIATION OF POISSON DISTRIBUTION

Successive r	Frequency f	Product rf	Product
0		0	0
1
2
3
…….	……..	……..	……..
r
……..	…….	……..	…….

Hence mean and variance of a Poisson distribution are equal to m. Similarly we can obtain,

MEAN DEVIATION

Show that in a Poisson distribution with unit mean, and the mean deviation about the mean is 2/e times the standard deviation.

Solution

  But mean = 1 i.e. m =1 and S.D. =

r	P (r)	|r-1|	P(r)|r-1|
0		1
1		0	0
2		1
3		2
4		3
…..	…..	…..	…..
r		r-1

Mean Deviation =

MOMENT GENERATING FUNCTION OF POISSON DISTRIBUTION

Solution

Let be the moment generating function then

CUMULANTS

The cumulant generating function is given by

Now cumulant =coefficient of in K (t) = m

i.e. , where r = 1,2,3,…

Mean =

RECURRENCE FORMULA FOR POISSON DISTRIBUTION

SOLUTION.

By Poisson distribution

On dividing (2) by (1) we get

Example. Assume that the probability of an individual coal miner being killed in a mine accident during a year is . Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.

Solution

Example. Suppose 3% of bolts made by a machine are defective, the defects occuring at random during production. If bolts are packaged 50 per box, find

(a)  Exact probability and

(b)  Poisson approximation to it, that a given box will contain 5 defectives.

Solution

(a)  Hence the probability for 5 defectives bolts in a lot of 50.

(b) To get Poisson approximation m = np =

Required Poisson approximation=

Example. In a certain factory producing cycle tyres, there is a smallchance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective, one defective and two defective tyres, respectively, in a consignment of 10,000 lots.

Solution.

S.No.	Probability of defective	Number of lots containing defective
1.
2.
3.

Normal Distribution-

The concept of normal distribution was given by English mathematician Abraham De Moivre in 1733 but the concrete theory was given by Karl Gauss that is why sometime normal distribution is called Gaussian distribution.

Normal distribution is a continuous distribution. It is a limiting case of binomial distribution.

The probability density function of a normal distribution is given by-

Here

Where

Note-

1. If a random variable X follows normal distribution with mean and variance then we can write it as- X

2. If X , then is called standard normal variate with mean 0 and standard deviation 1.

3. The probability density function of standard normal variate Z is given as-

Where

Graph of a normal probability function-

The curve look like bell-shaped curve. The top of the bell is exactly above the mean.

If the value of standard deviation is large then curve tends to flatten out and for small standard deviation it has sharp peak.

This is one of the most important probability distributions in statistical analysis.

Example:

1. If X then find the probability density function of X.

2. If X then find the probability density function of X.

Sol.

1. We are given X

Here

We know that-

Then the p.d.f.  will be-

2. . We are given X

Here

We know that-

Then the p.d.f.  will be-

Mean, median and mode of the normal distribution-

Let ‘a’ is the median, then it divides the total area into two parts-

Where-

Let a>mean, then-

Thus-

So that mean = median.

Note- mean deviation about mean is =

Mode-

The mode of the normal distribution is and modal ordinate is given by-

Hence the mean, median and mode are equal in normal distribution.

Area property of a normal distribution (Area under the normal curve)-

Let X follows the normal distribution with mean and variance

We form a normal curve by taking

Note- Total area under the curve is always 1.

Example: If a random variable X is normally distributed with mean 80 and standard deviation 5, then find-

1. P[X > 95]

2. P[X < 72]

3. P [85 < X <97]

[Note- use the table- area under the normal curve]

Sol.

The standard normal variate is –

Now-

1. X = 95,

So that-

2. X = 72,

So that-

3. X = 85,

X = 97,

So that-

Example: In a company the mean weight of 1000 employees is 60kg and standard deviation is 16kg.

Find the number of employees having their weights-

1. Less than 55kg.

2. More than 70kg.

3. Between 45kg and 65kg.

Sol. Suppose X be a normal variate = the weight of employees.

Here mean 60kg and S.D. = 16kg

X

Then we know that-

We get from the data,

Now-

1. For X = 55,

So that-

P[X < 55] = P[Z < - 0.31] = P[Z > 0.31]

= 0.5 – P [ 0 < Z < 0.31]

= 0.5 – 0.1217

2. For X = 70,

So that-

P[X > 70] = P[Z > 0.63]

= 0.5 – P[0 < Z < 0.63]

= 0.5 – 0.2357 = 0.2643

3. For X = 45,

For X = 65,

Hence the number of employees having weights between 45kg and 65kg-

Example: The mean inside diameter of a sample of 200 washers produced by a machine is 0.0502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

Sol.

Here-

And

Area for non-defective washers = area between z = -1.2 to +1.2

= 2 area between z = 0 and z = 1.2

= 2 × 0.3849 = 0.7698 = 76.98%

Then percent of defective washers = 100 – 76.98 = 23.02 %

Example: The life of electric bulbs is normally distributed with mean 8 months and standard deviation 2 months.

If 5000 electric bulbs are issued how many bulbs should be expected to need replacement after 12 months?

[Given that P (z ≥ 2) = 0. 0228]

Sol.

Here mean (μ) = 8 and standard deviation = 2

Number of bulbs = 5000

Total months (X) = 12

We know that-

Area (z ≥ 2) = 0.0228

Number of electric bulbs whose life is more than 12 months ( Z> 12)

= 5000 × 0.0228 = 114

Therefore replacement after 12 months = 5000 – 114 = 4886 electric bulbs.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Unit - 6

Probability

6.1 Random variable, discrete and continuous random variable, Probability density function

A random variable is a real-valued function whose domain is a set of possible outcomes of a random experiment and range is a sub-set of the set of real numbers and has the following properties:

i) Each particular value of the random variable can be assigned some probability

Ii) Uniting all the probabilities associated with all the different values of the random variable gives the value 1.

A random variable is denoted by X, Y, Z etc.

For example if a random experiment E consists of tossing a pair of dice, the sum X of the two numbers which turn up have the value 2,3,4,…12 depending on chance. Then X is a random variable 

Discrete random variable-

A random variable is said to be discrete if it has either a finite or a countable number of values. Countable number of values means the values which can be arranged in a sequence.

Note- if a random variable takes a finite set of values it is called discrete and if if a random variable takes an infinite number of uncountable values it is called continuous variable.

Discrete probability distributions-

Let X be a discrete variate which is the outcome of some experiments. If the probability that X takes the values of x is , then-

Where-

1.

2.

The set of values with their probabilities makes a discrete probability distribution of the discrete random variable X.

Probability distribution of a random variable X can be exhibited as follows-

X			and so on
P(x)

Example: Find the probability distribution of the number of heads when three coins are tossed simultaneously.

Sol.

Let be the number of heads in the toss of three coints

The sample space will be-

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Here variable X can take the values 0, 1, 2, 3 with the following probabilities-

P[X= 0] = P[TTT] = 1/8

P[X = 1] = P [HTT, THH, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P[HHH] = 1/8

Hence the probability distribution of X will be-

X
P(x)

Example: For the following probability distribution of a discrete random variable X,

Find-

1. The value of c.

2. P[1<x<4]

Sol,

We know that-

So that-

0 + c + c + 2c + 3c + c = 1

8c = 1

Then c = 1/8

Now, 2. P[1<x<4] = P[X = 2] + P[X = 3] = c + 2c = 3c = 3× 1/8 = 3/8

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-

P[X = 0, Y = 0] = 2/9 ,  P[X = 0, Y = 1] = 1/9,    P[X =1,  Y = 0] = 1/9

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-

Now-

But

So that-

Hence X and Y are not independent.

Continuous Random Variable:

A continuous random variable is a random variable where the data can take infinitely many values. For example, a random variable measuring the time taken for something to be done is continuous since there are an infinite number of possible times that can be taken.

Continuous random variable is called by a probability density function p (x), with given properties: p (x) ≥ 0 and the area between the x-axis  & the curve is 1: ... Standard deviation of a variable Random is defined by σ x = √Variance (x).

Continuous probability distribution (probability density function)-

When a random variable X takes every value in an interval, it gives to continuous distribution of X.

The probability distribution of a continuous variate x defined by a function f(x) such that the probabilities of the variate x falling in the interval

We express it as-

Here f(x) is called the probability density function.

Distribution function-

If F(x) = P(X≤x) = then F(x) is called as the distribution function of X.

It has the following properties-

1. F(-∞) = 0

2. F(∞) = 1

3.

Probability Density:

Probability density function (PDF) is a arithmetical appearance which gives a probability distribution for a discrete random variable as opposite to a continuous random variable. The difference among a discrete random variable is that we check an exact value of the variable. Like, the value for the variable, a stock worth, only goes two decimal points outside the decimal (Example 32.22), while a continuous variable have an countless number of values (Example 32.22564879…).

When the PDF is graphically characterized, the area under the curve will show the interval in which the variable will decline. The total area in this interval of the graph equals the probability of a discrete random variable happening. More exactly, since the absolute prospect of a continuous random variable taking on any exact value is zero owing to the endless set of possible values existing, the value of a PDF can be used to determine the likelihood of a random variable dropping within a exact range of values.

Probability density function-

Suppose f( x ) be a continuous function of x . Suppose the shaded region ABCD shown in the following figure represents the area bounded by y = f( x ), x –axis and the ordinates at the points x and x + δx , where δx is the length of the interval ( x , x + δx ).if δx is very-very small, then the curve AB will act as a line and hence the shaded region will be a rectangle whose area will be AD × DC this area = probability that X lies in the interval ( x, x +δx )

= P[ x≤ X ≤ x +δx ]

Hence,

Properties of Probability density function-

1. 

2.

Example: If a continuous random variable X has the following probability density function:

Then find-

1. P[0.2 < X < 0.5]

Sol.

Here f(x) is the probability density function, then-

Example. The probability density function of a variable X is

(i)                Find

(ii)             What will be e minimum value of k so that

Solution. (i) If X is a random variable then

(ii)Thus minimum value of k=1/30.

Example. A random variate X has the following probability function

x	0	1	2	3	4	5	6	7
P (x)	0	k	2k	2k	3k

(i)                Find the value of the k.

(ii)            

Solution. (i) If X is a random variable then

Example: Show that the following function can be defined as a density function and then find .

Sol.

Here

So that, the function can be defined as a density function.

Now.

Example:

Let X be a random variable with PDF given by

a, Find the constant c.

b. Find EX and Var (X).

c. Find P(X ).

Solution.

1. To find c, we can use

Thus we must have .

b.     To find EX we can write

In fact, we could have guessed EX = 0 because the PDF is symmetric around x = 0. To find Var (X) we have

c.      To find we can write

Example:

Let X be a continuous random variable with PDF given by

If , find the CDF of Y.

Solution. First we note that , we have

Thus,
 

Example:

Let X be a continuous random variable with PDF

Find .

Solution. We have

Unit - 6

Probability

6.1 Random variable, discrete and continuous random variable, Probability density function

A random variable is a real-valued function whose domain is a set of possible outcomes of a random experiment and range is a sub-set of the set of real numbers and has the following properties:

i) Each particular value of the random variable can be assigned some probability

Ii) Uniting all the probabilities associated with all the different values of the random variable gives the value 1.

A random variable is denoted by X, Y, Z etc.

For example if a random experiment E consists of tossing a pair of dice, the sum X of the two numbers which turn up have the value 2,3,4,…12 depending on chance. Then X is a random variable 

Discrete random variable-

A random variable is said to be discrete if it has either a finite or a countable number of values. Countable number of values means the values which can be arranged in a sequence.

Note- if a random variable takes a finite set of values it is called discrete and if if a random variable takes an infinite number of uncountable values it is called continuous variable.

Discrete probability distributions-

Let X be a discrete variate which is the outcome of some experiments. If the probability that X takes the values of x is , then-

Where-

1.

2.

The set of values with their probabilities makes a discrete probability distribution of the discrete random variable X.

Probability distribution of a random variable X can be exhibited as follows-

X			and so on
P(x)

Example: Find the probability distribution of the number of heads when three coins are tossed simultaneously.

Sol.

Let be the number of heads in the toss of three coints

The sample space will be-

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Here variable X can take the values 0, 1, 2, 3 with the following probabilities-

P[X= 0] = P[TTT] = 1/8

P[X = 1] = P [HTT, THH, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P[HHH] = 1/8

Hence the probability distribution of X will be-

X
P(x)

Example: For the following probability distribution of a discrete random variable X,

Find-

1. The value of c.

2. P[1<x<4]

Sol,

We know that-

So that-

0 + c + c + 2c + 3c + c = 1

8c = 1

Then c = 1/8

Now, 2. P[1<x<4] = P[X = 2] + P[X = 3] = c + 2c = 3c = 3× 1/8 = 3/8

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-

P[X = 0, Y = 0] = 2/9 ,  P[X = 0, Y = 1] = 1/9,    P[X =1,  Y = 0] = 1/9

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-