Probability Mass Function Examples

Introduction:

In probability, a probability mass function (pmf) is a function that gives the probability that a discrete random variable and accurately equals some value. Probability mass function examples differ from portable document format defined only for continuous random variables are not probabilities as such examples. The integral over a range of possible values [a, b] gives the probability of the random variable. Some examples for probability mass function are below. I like to share this Probability Set with you all through my article.

Example 1:

At the same time two coins are tossed, what is the probability result of getting (i) accurately one head (ii) no less than one head (iii) more or less one head.

Solution:

The sample space is S = {HH, HT, TH, TT}, n(S) = 4

Let A be the event of getting one head, B be the event of getting at least one

Head and C be the event of getting almost 1head.

∴ A = {HT, TH}, n(A) = 2

B = {HT, TH, HH}, n(B) = 3

C = {HT, TH, TT}, n(C) = 3

(i)               P(A) = n(A) / n(S) = 2/4 = 1/2

(ii)              P(B) =n(B)/n(S) = 3/4

(iii)             P(C) = n(C)/n(S) = 3/4


Example 2:


Two cards have chosen without replacement from 54 cards X measures the number of heart cards drawn y measures the number of clubs drawn. Find probability of joint probability mass function?

Solution:

A deck involves 52 cards.

Sample space = {all ordered pair (p, q)} = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0)}

Here, we have to find out the probability at each element of the sample space, and then we have to joint probability mass function.

X (0, 0) = (26 / 52) * (25 / 51) = 650 / 2652

X  (0,1) = (26 / 52) * (13 / 51) + (13 / 52) * (26 / 51) = 676 / 2652

X  (0,2) = (13/52) * (12/51) = 156 / 2652

X (1, 0) = (13/52) * (26/51) = 676/2652

X (1, 1) = (13/52) * (13/51) + (13/52) * (13/51) = 338/2652

X (2, 0) = (13/52) * (12/51) = 156/2652

X (s, t) = 0 if (s, t) is not in the sample space.

Free Calculus Tutoring

Introduction:

British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. Calculus has  two classifications: Differentiation calculus, Integration calculus. Calculus is the study of rates of change, area, or volume. In symbol, we require to find f(x) where, d/dx f(x) = g(x). Integration is one of most important study of calculus in mathematics. We require to find f(x) = ∫ g(x) dx.In the trigonometric calculus the derivative of a constant is zero. There is no exact value for the integral.


Examples for free calculus tutoring:

Differential free calculus tutoring:

Free calculus tutoring problem 1:

Find the first and second derivative of f(x) = 8x4 + 7x3 - 6x2 - 5x +3

Solution;

Steps to solve:

• First differentiate f(x) = 8x4 + 7x3 - 6x2 - 5x +3                                  
• we know, d / dx (x n ) = n xn-1
• when we differentiate the first term we get  (8×4) x (4-1)
• The above equation can above simplified as 32 x 3
• Like wise we can differentiate  and simplify all the terms in the equation
• Finally after the first derivative we will get  32 x 3 + 21 x2 -12x -5
• Now we will go for Second derivative
• Second derivative of  f(x) = 32 x 3 + 21 x2 -12x -5
• Differentiate all terms   (32 × 3)x2 + (21 × 2) x - (12 × 1) x0
• End of the second derivative we will get   96x2 + 42 x - 12

First derivative:

f' = df / dx

= (8 × 4) x (4-1) + (7 × 3)x(3-1) - (6 × 2)x(2-1) - (5 × 1)x (1-1)

= 32 x 3 + 21 x2 – 12x -5

Second derivative:

f '' = df ' / dx

= 32 x 3 + 21 x2 – 12x -5

= 96x2 + 42 x - 12

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 ` (8x4 + 7x3 - 6x2 - 5x +3) = 96x2 + 42 x - 12



Free calculus tutoring problem  2:

Find the first and second derivative of sin 2x.

Solution;

Steps to solve:

• First differentiate f(x) = sin 2x                                                       
• (we know, d/dx (sin x) = cos x)
• when we differentiate sin x  = cos x So, we get  sin 2x = cos 2x
• Then we differentiate 2x  So, we get  2
• The differentiation of sin 2x = 2 cos 2x
• Finally after the first derivative we will get  2 cos 2x
• Now we will go for Second derivative
• we know,  d/dx (cos x) = −sin x
• The second derivative 2 cos 2x = 2 ×2 (- sin 2x)
• So, The second derivative of  sin 2x = -4 sin 2x

First derivative:

f' = df / dx

= `d/dx` (sin 2x)

= 2 cos 2x

Second derivative:

f '' = df ' / dx

= `d /dx` (2 cos 2x)

= -4 sin 2x

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 `(sin 2x) = -4 sin 2x

I like to share this calculus problems with you all through my article.

Integral free calculus tutoring:


Free calculus tutoring problem 1:

Integrate  `int ` (-2x2 + 8x3 + 2x4 + x5) dx

Solution;

Steps to solve:

• First seperate each term ` int` (- 2x2 + 8x3 + 2x4 + x5) dx
• So, we get `int` (- 2x2 +8x3 + 2x4 + x5) dx =  `int ` - 2x2 dx + ` int ` 8x3 dx + `int` 2x4 dx + `int`   x5  dx
• Integrate first term ` int ` - 2x2 dx =  -2 (1/3) x3                                                                  
•  [ we know,   `int` xn dx = 1/ n+1 (xn+1)]
• Integrate all term and we get   (- 2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6
• Finally after the integration we will get  (- 2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

` int ` (-2x2 + 8x3 + 2x4 + x5) dx =` int` -2x2 dx + `int` 8x3 dx + ` int ` 2x4 dx +` int` x5  dx

= -2 `int ` x2 dx +  8  `int` x3 dx + 2` int`x4 dx + `int` x5  dx

= -2 (1/3) x3 + 8(1/4) x4 + 2(1/5) x5+(1/6) x6

= (-2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6

= (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

Answer for the given free calculus tutoring problem is:

` int` (-2x2 + 8x3 + 2x4 + x5) dx = (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

How to solve linear programming problems

Linear Programming is one of the operations research techniques. It is one of the best mathematical techniques for finding the limited use of resources of a concern in a best way. Complex problems can be modeled using linear functions in a presentable way by the management. The linear programming technique is used in solving a wide range of operations management problems.

Definition of linear programming problems:

Linear Programming is defined as a technique which allocates the available resources in an optimum manner for achieving the company’s objective which is for maximizing the overall profit or to minimize the overall cost under conditions of certainty.

Linear Programming can be applied to areas which are given below:

Allocation of resources to various activities of the concern, for example: man power, machine etc.
Production scheduling.
The common characteristics in the above mentioned areas are to allocate limited resources to the activities of the concern.

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How to solve Linear Programming Problems: Mathematical Formulation

Linear Programming can be used in a variety of situations. In most of the business or economic situations, the resources will be limited, the problem there will be to make use of the available resources in such a way as to maximize the production or to maximize the profit or to minimize the expenditure. This can be formulated as linear programming models.

Mathematical Formulation of the problem:

How to solve linear programming problems?? here are the steps which you need to follow:

Step 1:

Write down the decision variables of the problem.

Step 2:

Formulate the objective function to be optimized as a linear function of the decision variables.

Step 3:

Formulate the other conditions of the problem as Linear equations or In equations in terms of the decision variables.

Step 4:

Add the non negativity constraint from the consideration that negative values of the decision variables do not have any valid physical interpretation.

The objective function, the set of constraints, and the non negative constraints together form an LPP.


Steps to solve linear programming problems using Graphical Method:


When a LPP has only two variables in the objective function and constraints, it can be easily solved using the graphical method. The given information of a LPP can be plotted on the graph and the optimal solution can be obtained from the graph.

The steps to solve an Linear Programming Problem using Graphical method is given below:

Step 1:

Identify the decision variables, the objective function and the restrictions for the given Linear Programming Problem (LPP).

Step 2:

Write the Mathematical Formulation of the problem.

Step 3:

Plot the points on the graph representing all the constraints of the problem. Find the feasible region or solution space. The intersection of all the regions represented by the constraints of the problem is called the feasible region and is restricted to the first quadrant only.

Step 4:

The Feasible region obtained in the step 3 may be bounded or un bounded. Determine the Co-ordinates (x, y) values of all the corner points of the feasible region.

Step 5:

Find the value of the objective function at each corner points (solution) determined in step 3.

Step 6:

Select a point from all the corner points that optimizes (Maximizes or Minimizes) the values of the objective function. It gives the Optimum Feasible Solution.

Understanding graphing systems of linear equations is always challenging for me but thanks to all math help websites to help me out.

Some Exceptional Cases of Linear Programming Problem:


There may be an LPP for which no solution exists or for which the only solution obtained is an unbounded one. The exceptional cases arise in the application of graphical method are

• Alternative Optima
• Unbounded Solution
• Infeasible Solution or Non existing Solution

Alternative Optima:

When the objective function is parallel to the binding constraint, the objective function will assume the same optimal value at more than one solution point, because of this reason, they are called as Alternative Optima.

Unbounded Solution:

When the values of the decision variables may be increased in definitely without violating any of the constraints, the feasible region is unbounded. In such cases, the value of the objective function may increase (for maximisation) or decrease (for minimisation) in definitely. Thus, both the solution space and the objective function value are unbounded.

Infeasible Solution:

When the constraints are not satisfied simultaneously, the LPP has no feasible solution. This solution can never occur, if all the constraints are of less than or equal to type.




Example for some exceptional cases:


The general form of the LPP is used to develop the procedure for solving a common programming problem.

A standard LPP Some exceptional cases is of the form
Max (or min) Z = c1x1 + c2x2 + … +cnxn
x1, x2, ....xn these are called decision variable.

Ex: Show graphically that the model

Maximize Z = -5y

Subject to

x+y<span style="font-family: Serif;">?</span> 1

0.5x-5y<span style="font-family: Serif;">?</span> -10

x<span style="font-family: Serif;">?</span> 0

y<span style="font-family: Serif;">?</span> 0 has no feasible solution.

Sol:

Draw the graphs x + y = 1

- 0.5 -5y = - 10

Shade the half planes of the constraints x + y 1 …(1)

-0.5x - 5y -10 …(2)



Points are (0,1)(0,2)(1,0)(20,0)

Note that the origin (0, 0) does not satisfy the in 2nd equation hence the required region is the upper half plane.

From the graph, that the intersection of the constraints is empty. Therefore the given problem has no feasible solution. So, the some exceptional cases of given LPP has no solution.

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slope formula calculator

• If an area of surface tends evenly towards top or down, it is referred as slope.

• The slope of a line is usually denoted by m.

• In other words, slope is the ratio of change in the y coordinates to the change in the x coordinate. The slope is otherwise named as gradient. Slope is equal to rise divided by run.

• In general, mathematical calculators are used to perform mathematical operations .In this article of slope formula calculator, we are going to learn how to find the slope between points using the calculator.

I like to share this slope of the line with you all through my article.

How to calculate Slope formula:


If the two points (x1, y1), (x2, y2) are given, the slope formula is given by

Slope   m = (y2-y1) / (x2-x1)

Step by step explanation:

The steps necessary for finding slope between two points using calculator are given below:

Step 1:  Enter the x1 and x2 values.

Step 2:  Enter the y1 and y2 values.

Step 3: The slope between two points will be shown in the result area.

Having problem with Negative Slope Read my upcoming post, i will try to help you.

Example calculation using slope formula:


1) Using the slope formula, Find the slope of the line through the points (6, 10) and (9, 11)

Solution:

Given:            x1 = 6     x2 = 9

y1 = 10    y2 = 11

Slope   m = (y2-y1) / (x2-x1)

= (11 -10) / (9 - 6)

= 1/3

2) Find the slope of the line through the points (7, 11) and (12, 14)

Solution:

Given:            x1 = 7     x2 = 12

y1 = 12    y2 = 14

Slope   m = (y2-y1) / (x2-x1)

= (14 -11) / (12 - 7)

= 3/5

3) Find the slope of the line through the points (11, 14) and (16, 23)

Solution:

Given:            x1 = 11     x2 = 16

y1 = 14    y2 = 23

Slope   m = (y2-y1) / (x2-x1)

= (23 -14) / (16 - 11)

= 9/5

4) Find the slope of the line through the points (21, 25) and (31, 35)

Solution:

Given:            x1 = 21     x2 = 31

y1 = 25    y2 = 35

Slope   m = (y2-y1) / (x2-x1)

= (35 -25) / (31 - 21)

= 10/10

= 1.


Practice problems on slope formula calculator:


1) Find the slope of the line through the points (11, 10) and (12, 11)

Answer: m = 1

2) Find the slope of the line through the points (17, 15) and (23, 18)

Answer: m = 1/2

Solving Online Type of Quadrilaterals

Solving online problems is used for learning problems through online that help students to learn easy and transfer knowledge and skills to people through online. Learn online will help kids to study anywhere at anytime.Quadrilaterals are four sided polygons. They are classified by their sides and angles. an important distinction between quadrilaterals is whether or not one or more pairs of sides are parallel. One of the more familiar quadrilaterals is a parallelogram. We see that a square, a rectangle, and a rhombus are all different types of a parallelogram. The quadrilaterals are of 4 types basically but there are some other types that satisfy the properties of quadrilaterals. Let us see about solving online type of quadrilaterals.



Solving Online Type of Quadrilaterals:

Here let us see type of quadrilaterals and its properties,

Trapezoid:

A trapezoid is a quadrilateral that has one pair of parallel sides.



Parallelogram:

A parallelogram is a quadrilateral of  two pairs of parallel sides.



Additional properties:

Opposite sides parallel
Opposite sides equal in measure
Opposite angles equal in measure

Understanding Area of a Parallelogram is always challenging for me but thanks to all math help websites to help me out.

Rectangle:

A rectangle is a parallelogram with four right angles.


Additional properties:

Opposite sides parallel
Opposite sides equal in measure
All angles measure 90°
Diagonals equal in length

Square:

A square is a rectangle with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
All angles measure 90°
Diagonals equal in length

Rhombus:

A rhombus is a parallelogram with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
Opposite angles equal in measure

Isosceles Trapezoid:

An isosceles trapezoid is a quadrilateral.


Additional properties:

One pair of parallel sides
Nonparallel sides are equal in length
Solving Online Type of Quadrilaterals:

Practice problems for solving online type of quadrilaterals,

Example 1:

Find the base of a parallelogram if its area is 512 cm2 and altitude is 14 cm.

Solution:

Area = base × height.

512 = base × 14.

b = 512 / 14

= 512 cm.

Base = 36.5 cm.

Example 2:

Find the perimeter of square whose sides are 11 cm.

Solution:

given the side if square is 11cm

Perimeter of the square, P = 4a

= 4 × 11 cm

= 44 cm

Hence the perimeter of square is 44 cm.

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Squeeze Theorem Proof

The proof is a correct demonstration of math statement. It should be true. The proof of one statement is used in other statement proof. The limits of a function are defined by squeeze theorem. We can also refer the squeeze theorem as sandwich theorem or pinching theorem. Now we are going to see about squeeze theorem.

Explanation for Squeeze Theorem Proof

Define squeeze theorem:

In calculus, the squeeze theorem is used and we can analyze the function’s limits by using this theorem. The interval ‘I’ is including the point ‘a’. The functions f, g and h are derived by interval ‘I’. We gets the function as g(x) `lt=` f(x) `lt=` h(x) if we have unequal 'x' values present. In other words, the definition is `lim_(x->a)` g(x) = `lim_(x->a)` h(x) = L. Final result is `lim_(x->a)` f(x). The lower bound g(x) and upper bound h(x) are used to bounds the f(x). We does not including the value 'a' in the interval 'I'.

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More about Squeeze Theorem Proof

The squeeze theorem proof is derived by previous statements and the proof of this theorem is using the special case and general notations.

Using the special cased for squeeze theorem proof:

The special case is g(x) = 0 for all x and L = 0.

First prove the special case as,

Let us take the special case as `lim_(x->a)` h(x) = 0.

Use the fixed positive number ε > 0 and δ > 0 from limits of function.

if 0 < | x-a | < δ then |h(x)| < ε.

Take the terms from above interval and the terms are 0 = g(x) `<=` f(x) `<=` h(x). So | f(x) | `<=` | h(x) |.

The conclusion of theorem is if 0 < | x-a | < δ then | f(x) | `<=` | h(x) | `<` ε.

Final proof of squeeze theorem by above condition is `lim_(x->a)` f(x) = 0 = L.

Use the general notations for squeeze theorem proof:

The g and L are arbitrary and these are used for proof. We have g(x) `<=` f(x) `<=` h(x).

Subtract the g(x) from both sides as 0 `<=` f(x) - g(x) `<=` h(x) - g(x).

Let us take x--> a and g(x) and h(x) as 'L'.

Therefore, h(x) - g(x) --> L - L = 0.

The special case is used in theorem conclusion as f(x) = (f(x) - g(x)) + g(x) --> 0 + L = L.

Hence the sandwich theorem is proved.

Metric Calculator

Metric number line is one of the important topics on the metric number system in mathematics subject. The quantities used to find lengths, capacities, weights of things etc are called measures. Many Countries have their own system of measures. But Metric System of measures is very simple and easy To calculate. Hence most countries in the world use Metric System of measures.

In Metric System,

• Length’s basic unit is metre(m)
• Weight’s basic unit is gram(g)
• Capacity‘s basic unit is litre(l)

And Another name of metric system is decimal system.

I like to share this Metric Unit Converter with you all through my article.

Explanations of the Metric Number Line:

• The metric system is used in all the places of the world, because of its superior basis metric units are linked to each other by factor of 10.

• So while we convert one metric unit to another, we have to move the decimal point in the unique value.

• These metric number lines give us an easy way to do these unit conversions.

Metric units Names and Abbreviations are given below:

• Centi-meter = cm
• Milli-meter = mm
• Kilo-meter = km
• Mega meter = Mm
• Decimeter = dm
• Dekameter = dam
• Hectometer = hm
• Micrometer = mcm
• Meter = m
• Liter = L
• Grams = g
• Volt = v

The above names of units are important. And those all units are different like meter is length, liter is volume, gram is mass or weight.

Metric Number Line Units:

• In number line we can forever multiply the unit by a factor of 10.
• Let us make the number line and use unit as the basic unit.
• Then we can use the metric number line for all units.
• The unit could be decigrams, deciliters, decimeters, decivolts, etc…

Example figure for metric number line:



Ex :  Convert dekagrams (dag) to centigrams (cg) :



The above figure represents the example of converting metric numbers from line, here we have to choose the start point to end with stop point. Here we note two things, there is

1. Direction and,

2. The number of points necessary to move to get the stop point.

Then we have to make the decimal changes into the original number consequently.

For the above number line, we have to move three places to the right direction.



For ex it can be 4.5 dekagram means; it could be changed into 4500 cg. Because of moved three places to right direction.

Having problem with What is Geometric Mean Read my upcoming post, i will try to help you.
 

These all are the metric number line details.

Metric Calculator - Conversion Tables and Examples:

In decimal system  to concert a higher value into lower value then we need to multiply it by powers of ten and to convert lower value into a higher value then we need to divide the number by ten.

Let us discuss about linear measure  like measures of length,weight,capacity.

Length-Conversion Table:

10mm	1cm
10cm	1dm
10dm	1m
10m	1dam
10dam	1hm
10hm	1km
100cm	1m
1000m	1km

Weight - Conversion Table:

10mg	1cg
10cg	1dg
10dg	1g
10g	1dag
10dag	1hg
10hg	1kg
1000mg	1g
1000g	1kg
100kg	1quintal(q)
1000kg	1 tonne(ton)

Capacity / Volume :- Conversion Table:

10ml	1cl
10cl	1dl
10dl	1l
10l	1dal
10dal	1hl
10hl	1kl

Ex 1 : Convert 4 km into lower units.

Sol:

4 km = 40 hm (4 × 10) = 4 × 101 hm

= 400 dam (4 × 100) = 4 × 102 dam

= 4000 m (4 × 1000) = 4 × 103 m

=40000 dm (4 × 10000) = 4 × 104 dm

= 400000 cm (4 × 100000) = 4 × 105 cm

= 4000000 mm (4 × 1000000) = 4 × 106 mm

Ex 2: Express 1267547 mm into higher units.

Sol:

12675477 mm   = 126754.7cm[1267547/10]  =1267547 x 10-1 cm

=12675.47dm[1267547/100] =1267547 x 10-2 dm

= 1267.547m[1267547/1000] =1267547 x 10-3 m

=126.7547dam[12.6754]/10000  = 1267547 x 104 dam

=12.67547hm[1267547/100000]  =1267547 x 10-5 hm

=1.267547km[1267547/1000000] =1267547 x 10-6 km

Ex 3: Convert 7m into millimeter.

Sol:

We know that ,

1m =1000mm

Therefore ,7m  = 7*1000mm

=7000mm

Ex 4: Express 10kg5dag in grams

Sol:

We know that,

1kg   =1000g

1dag =10g

Therefore, 10kg5dag = 10 *1000g +5 *10g

=10000g +50g

=10050g

Understanding very hard math problems is always challenging for me but thanks to all math help websites to help me out.

Practice problem to help with metric calculation:

1)Convert 6m into millimetre.

Ans: 6000mm

2)Express 2769 g in kilograms.

Ans: 2.769 kg

3)Convert 25 kl 37 l into litres.

Ans: 25037 l