Three Types of Probability

Introduction to three types of probability:

Three types of probability article deals with the definition of three types of probability and the model problems related to it.

Types of probability:

The probability is classified in to

Probability as counting.
probability as geometry
Probability as algebra, are the three types and the explanation is given below

Three Types of Probability:

Probability as counting:

This probability is very simple and easy to understand. This is used to find the probability of number of times an event occurs. This probability is calculated by dividing the number of successive out come by the total number of outcomes.

Probability as counting =` (("number of sucessive outcome")/ ("total number of outcomes"))`

Probability as geometry:

This probability is suitable for one or more continuous variable. We can find the probability dividing the length of successive part by the length of total parts in the event.

Probability as geometry = `(("length of successive part")/ ("length of total part"))` .

Probability as algebra:

This is applicable to problems where we can find the enough relationship between two probabilities, at that time we can solve the problem algebraically.Having problem with 10 cbse sample papers keep reading my upcoming posts, i will try to help you.

Model Problems to three Types of Probability:

Problem 1: what is the probability of getting 5, when a dice is thrown?

Solution:

The sample space S = {1, 2, 3, 4, 5, 6}

n(S) = 6

Getting 5 = {5}

n (5)= 1

Probability of getting 5 = `(n (5))/(n(S))`

= `(1/6)`

Probability of getting 5 is `(1/6)` .

Problem 2:

The probability of getting mark in next test is x, which is likely, equal to three the marks he will not get.what is the probability of marks in next test.

Solution:

Getting mark in next test is x

Probability he will not get is 1-x

By given data:

x =2(1-x)

x= 2-2x

x+2x=2

3x =2

x=`2/3`

The probability of getting mark in next test is `2/3`

Problem 3:

John is working from 8am to 5pm, every 15minutes he will go out. What is the probability of going out from 4pm to 5pm?

Solution:

John goes out every 15 minutes.

Probability of going out from 4pm to 5pm = `(15minutes)/(60 minutes)`

= `(15/60)`

= `(1/4)`

Probability of going out from 4pm to 5pm is `(1/4)`

Sum of Poisson Random Variables

Introduction to sum of Poisson random variables:
The sum of the Poisson random variables is the mean value for the Poisson distribution. Poisson distribution is the limited form of the binomial distribution. The sum (mean) is calculated by using the number of the possibilities and the probability for the occurrence of the event. The example for the Poisson distribution is the number of cars passing through the certain street in the time period of t. This article has the information about the sum of the Poisson random variables.

Formula for the Sum of the Poisson Random Variables:

The formula used to find the sum (mean) of the Poisson distribution is

`lambda` = n p

Where n is the number of trial for the event and the p is the probability of the success.

Examples for the Sum of the Poisson Random Variables:
Examples 1 for the sum of the Poisson random variables:

Two coins are tossed for 33 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 33. So the value of n is 33.

The probability for getting one head with one coin is `1/2` .

The probability for getting two head with two coins are p = `(1/2) ^2` = `1/4` .

The sum (mean) `lambda ` = n p

`lambda ` = 33 (1/4)

` lambda` = `33 /4`

` lambda ` = 8.25

The sum (mean) for getting the head for the two coins is 8.25.Having problem with practice pre algebra keep reading my upcoming posts, i will try to help you.

Examples 2 for the sum of the Poisson random variables:

Four coins are tossed for 122 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 122. So the value of n is 122.

The probability for getting one head with one coin is `1/2` .

The probability for getting four head with four coins are p = `(1/2) ^4` = `1/16` .

The sum (mean) `lambda` = n p

`lambda` = `122 (1/16)`

`lambda` = `122 /16`

`lambda` = 7.625

The sum (mean) for getting the head for the four coins is 7.625.

Examples 3 for the sum of the Poisson random variables:

Five coins are tossed for 150 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 150. So the value of n is 150.

The probability for getting one head with one coin is `1/2` .

The probability for getting five head with five coins are p = `(1/2) ^5` = `1/32` .

The sum (mean) `lambda` = n p

`lambda` = `150 (1/32)`

`lambda ` = `150 /32`

`lambda` = 4.6875

The sum (mean) for getting the head for the five coins is 4.6875.

Vertices of a Hyperbola

Introduction to vertices of a hyperbola:

A hyperbola is defined as a smooth curve which tends to lie down in the plane and it can be described by the geometric properties or by the type of equation for which it may be the solution set. A hyperbola can be defined in the form of a curve that is represented by the function f(x) = 1 / x in the Cartesian plane.

In this article we are going to see about the vertices of a hyperbola.

How to Find the Vertices of a Hyperbola:

In the case of ellipse, there exists a relationship between x, y, and z axes with ellipses, the computations are long and painful. So that, for hyperbolas (where x < z), the relationship is z2 – x2 = y2 or, which means the same thing, z2 = y2 + x2.  This has been proved using the Pythagorean Theorem.
When the transverse axis is in the horizontal plane (when the center, and vertices line up side by side which is parallel to the horizontal axis), then the x2 goes with the part of the hyperbola's equation, and the y part is subtracted.
((a – y) 2 / x2) – ((b – c) 2 / y2) = 1.

Example to Find the Vertices of a Hyperbola:

Example 1: Given a hyperbola (x2 / 16) – (y2 / 4) = 1. Find the vertices of the given hyperbola.

Solution:

The given hyperbola opens right or left because it is of the form x – y

Here a2 = 16, b2 = 4, c2 = 16+4 = 20.

Hence a = 4, b = 2 and c = 4.5

The vertices of a hyperbola are given by (a, 0), (-a, 0)

Thus the vertices are (4,0), (-4,0).

Example2: Given a hyperbola (y2 / 25) – (x2 / 9) = 1. Find the vertices and foci of the given hyperbola.

Solution:

The hyperbola opens up and down because it is of the form y – x.

Here a2 = 25, b2 = 9, c2 = 25 + 9 = 34.

a = 5, b = 3, c = 5.8

The vertices of a hyperbola are given by (0, a), (0,-a)

Vertices of a hyperbola: (0, 5) and (0,-5)

The foci of a hyperbola are given by (0, c), (0,-c)

Foci of the hyperbola: (0, 5.8), (0,-5.8)

Acute Triangle Calculator

Introduction about triangle:
A triangle is three side polygons. Triangle is one of the important shapes in geometry.

Types of triangle

Equilateral triangle
Scalene triangle
Right triangle
Isosceles triangle
Acute triangle
Obtuse triangle
In this article we shall discuss about the acute triangle. A triangle has three sides and three angles. A triangle angles are less than 90 degree, such an angle is called acute triangle.

Acute means the angle should be less than 90 degree.

To find area of the acute triangle = `1/2` * base * height

Acute Triangle Calculator:




In calculator you have to enter the value of height and base of the triangle after pressing the calculate button area of the triangle is calculated and display on answer box.

Problem 1:

Calculate the area of the acute triangle whose base is 10cm and height is 12 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base =10 cm Height = 12 cm

Area = `1/2` * 12 *10

= 60 cm2

The answer is 60 cm2

Problem 2:

Calculate the area of the acute triangle whose base is 8cm and height is 11 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base = 8 cm Height = 11 cm

Area = `1/2` * 8 *11

= 44 cm2

The answer is 44 cm2

Problem 3:

Calculate the base of the acute triangle whose area is 121cm2 and height is 6 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Area =144 cm2 Height = 6 cm

Area = `1/2` * b *16

121 = `1/2` * b *16

B=121*`2/16`

B= 15.13 cm

The answer is 15.13 cm

Problem 4:

Calculate the height of the acute triangle whose base is 12cm and area is 196 cm2

Solution:

Area of the acute triangle = `1/2` * base * height

Base =12 cm Area = 196 cm2

Area = `1/2` * 12 *h

196 = `1/2` * 12 *h

h=196*`2/10`

B=39.2 cm

The answer is 39.2cm

Normal Sum Distribution

Introduction to normal sum distribution:

Let us see about normal sum distribution,

The normal distribution is also known as the normal probability distribution, happens to be the mainly helpful theoretical distribution for continuous variables. A lot of statistical data sum regarding business and economic problems are display in the type of normal sum distribution.In fact normal distribution is known as the ‘corner stone’ of  Modern statistics.Just like the Poisson distribution, the normal sum  distribution  may also be regarded as a limiting case of binomial distribution.

Definition of Normal Sum Distribution:

Let we see about definition of normal sum distribution,

A variable X is told that a normal distribution with parameter µ and s (or µ and s2) if the probability function is,

Here, Z =(X – µ) s

Constants of Normal distribution:

Mean = µ

Variance = s2

Standard deviation = s

The graph sum of the normal curve is shown below,

The distribution with µ = 0 and s2 = 1 is known as the standard normal distribution.

Understanding problem solving with proportions is always challenging for me but thanks to all math help websites to help me out.

Normal Sum Distribution-example Problems:

1) If X is normally distributed having mean 7 and standard deviation 6 find. P(0 = X = 10).

Solution :

Given, µ = 7, s = 6

(i) P(0 = X = 10)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 7) / 6 = -1.16

When X = 10, Z =(10 – 6) / 6 = 4/6 = 0.67

? P(0 = X = 10) = P(-1.2 < Z < 0.67)

= P( 0 < Z < 1.2) + P( 0 < Z < 0 .67) (due to symmetry)

= 0.3849 + 0.258

= 0.516

2) If X is normally distributed having mean 10 and standard deviation 8 find. P(0 = X = 20).

Solution :

Given, µ = 10, s = 8

Find, P(0 = X = 20)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 10) / 8 = -1.3

When X = 10, Z =(20 – 10) / 8= 10/6 = 1.3

? P(0 = X = 20) = P(-1.3 < Z < 0.3)

= P( 0 < Z < 1.3) + P( 0 < Z < 0 .3) (due to symmetry)

= 0.4032 + 0.4032

= 0.8064

Constant Rate of Change Math

Introduction to constant rate of change math:

Rate on which reliant output modify by value to vary within independent input is describing the rate of change. Functional association be usually indicated through y = f(x). Alteration within y is indicated through ?y also alteration within x is indicated through ?x. Consequently, as of the function y = f(x), we contain y + ?y = f(x + ?x).  In addition to constant rates of alteration are known through method, ?y / ?x.

Constant Rate of Change in Math:

In math consider the function is the linear function that is graph of y also x is a line after that the function is known through y = mx + x. In linear function rates of change is a constant which is known through its slope, m. Slope of the line can be find out through the following

m = ?y / ?x =  (y2 – y1/(x2 – x1)       

In math consider the function is non-linear then its rates of alteration are dissimilar on dissimilar points. Therefore, for such functions derived are estimated through distinguish the function toward get the rates of modify. Derivative of y through value to x can be written as dy/dx.

Examples for Constant Rate of Change Math:

Example 1

Solve the rate of change for the line to exceeds through points (5, 4) and (7, 10)

Solution:

Here, x1= 5, x2=4, y1=7 and y2=10

so, slope m = (y2 -  y1)/(x2- x1)

m = (10 - 4)/(7 - 5) = 6/2 = 3

Consequently, rate of alteration otherwise slope of line is 3.

Example 2

Solve the rate of change of function, f(x) =4x2 + 10x at x = 6

Solution:

f(x) =4 x2 + 10x

Derivative f '(x) = 8x + 10

Substitute x = 6 in f '(x) which gives,

f '(6) = 8*6 + 10 = 48 + 10 = 58

Therefore, rate of change of function f(x) at x = 6 is 58.

Example 3

Solve the rate of change of function, f(x) = 3x3 + 3x2 - 7x at x = 3

Solution:

f(x) =3x3 + 3x2 - 7x

Derivative f '(x) = 9x2+6x -7

Substitute x = 3 in f '(x) which gives,

f '(6) = 9(32)+6(3) - 7 = 81 + 18-7 = 92

Therefore, rate of change of function f(x) at x = 3 is 92.

Precalculus Math Final Help

Introduction to Precalculus Math Final Help:

The Precalculus is one of the foundation class of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus math final help. The following are the examples involved in Precalculus math final help.

Precalculus Math Final Help Problems:

Example 1:  Express the real and imaginary parts for 28 - i `sqrt(14) `

Solution:        Let z= 28 - i `"sqrt(14)`      

Re(z) = 28     Im(z) = `sqrt(14)`

Example 2: For what value of m the vectors `veca` and `vecb` are perpendicular to each other

`veca` =m`veci` + `vecj`` + ``veck`   `vecb`= 7`veci` - 9 `vecj` + 2 `vecj`

Solution: Given `veca` `_|_``vecb `

So, `veca` `vecb` = 0

=> (m`veci` +  `vecj`` + ``veck`) . (7`veci` - 9 `vecj``veck` )

=>  7m - 9 + 2

=>  7m  = 7

m =`7/7`  =1

Example 3: Solve for x,  ln(x-3) + ln(x-2) = ln(2x -6 )

Solution:   Simplify left side, So

ln(x-3) + ln(x-2)= ln(x - 3) (x - 2)

Therefore the given equation can be rewritten as

ln(x - 3)(x - 2) =  ln (2x - 6)

Taking exponent on above equation

eln(x - 3)(x- 2) = eln(2x - 6)

Simplifying the above equation

(x - 3) (x - 2)  = (2x - 6)

x2  - 5x + 6 = 2x - 6

x2  - 7x +12 = 0

Factoring the above equation, we get

(x - 4) (x - 3) = 0

When products of two factors are equal to 0, then one of the factors must be equal to 0.

So, x-4 = 0 => x=4

x-3 = 0 => x=3

Here x = 4 & 3 is the solution.

Example 4:   If f(x) = 14x-1, find f-1(y)?

Solution:Let y = 14x - 1

The given equation can be rewritten as

y + 1 =14x

=> x = y + 1
14

Therefore the f-1(y) = y + 1
14

Example 5: Write the given number in complex form `sqrt(-69)`

Solution: `sqrt(-69)` = `sqrt((-1)(69))`

= `sqrt(-1)` * `sqrt(69)`

= i `sqrt(69)`

Precalculus Math Final Help Practice Problems:

Problem 1:Write the given number in complex form `sqrt(-1415)`

Answer: = i `sqrt(1415)`

Problem 2:Write the real and imaginary parts for 89 - i `sqrt(75) `

Answer:   Re(z) = 89     Im(z) =`sqrt(75)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9