General Probability

General probability is the part of mathematics that learns the feasible outcomes of given events together with the outcomes' relative possibilities and distributions. In general usage, the word "probability" is used to mean the opportunity that a particular event will occur expressed on a linear scale from 0 to 1, also expressed as a percentage between 0 and 100.Now we will see the examples of probability.

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Examples- General Probability

Example 1

In a class there are 8 students got top eight marks in English. The marks are 78,83,86,88,92,93,95,98.  What is the probability for the following outcomes?

i) Select the marks are below 85.

ii) Select the marks between 80 and 90.

Solution:

i) Take P(A) is the probability for the marks are below 85.

Given marks are 78,83,86,88,92,93,95,98.

Total numbers n(S)=8

Here the following marks are below 85 n(A)={78,83}=2

So P(A)=(n(A))/(n(S))

=2/8

=1/4 .

ii) Take P(B) is the probability for the marks between 80 and 90.

The marks 83,86,88  are available between the 80 and 90.So n(B)=3

Total outcomes n(S)=8

So P(B)=(n(B))/(n(S))

= 3/8 .

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Example 2

What is the probability for select the letter ‘G’ from the word ‘GENERAL KNOWLEDGE’?

Solution:

Given word is GENERAL KNOWLEDGE.

Total letters n(S)=16

Number of ‘G’ letter n(A)=2

So the probability=2/16

=1/8 .

Example 3

Paul has 20 caps and Alex has 15 caps. What is the probability for select the Alex’s caps?

Solution:

Paul’s caps n(A)=20

Alex’s caps n(B)=15

Total number of caps n(S)=20+15

=35

Probability for select the Paul’s caps= 15/35

= 3/7 .
Practice Problems- General Probability

1) What is the probability for select the letter ‘E’ from the word ‘ELEMENT’?

Answer:

Probability=3/7 .

2) Tom has 5 white balls, Joseph has 7 yellow balls. What is the probability for getting the yellow balls?

Answer:

Probability=7/12 .

These examples and practice problems are used to study the general probability.

Solving Summation Notation

A summation notation (Σ) is used for sum the numbers or quantities indicated. We can sum all types of numbers such as real, integer, complex, rational numbers using sigma notation. Summation notation is also called as sigma notation. Using summation notation we can solve the problems easily. Summation a concise and convenient way of writing long sums.


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Solving Summation Notation – Solving Example Problems

Example 1: Find the sum for the given expression and explain the terms of summation notation;

7

∑ (n + 1)3 =?

n=1

Solution:

7

∑ (n + 1)3 =?

n=1

Lower Bound: ‘n’ is the lower bound of the summation notation.

Upper Bound:  ‘7’ is the upper bound of the summation notation.

Formula: (n + 1)3 is the formula to describe the each term of the summation notation.

Start: n = 1, here 1 is indicating the first number of the terms of summation notation. Usually value of start is zero (0) or one (1).

End: 7 is indicating the last umber of the terms of the summation notation.

Index Variable: Index variable ‘n’ is used to labeled each term of summation notation.

7

∑ (n + 1)3 = 8 + 27 + 64 + 125 + 216 + 343 + 512 = 1295

n=1

7

Therefore, ∑ i3 = 1295

i=1

Example 2: Solve and find the sum for the given expression;

6

∑ 3(n + n2) =?

n=1

Solution:

6

∑ 3(n + n2) =?

n=1

Formula: 3(n + n2), starting term: 1, end term: 6, and index variable: n

6

∑ 3(n + n2) / n = 3(1 + 12) + 3(2 + 22) + 3(3 + 32) + 3(4 + 42) + 3(5 + 52) + 3(6 + 62)

i=1


= 3(1 + 1) + 3(2 + 4) + 3(3 + 9) + 3(4 + 16) + 3(5 + 25) + 3(6 + 36)

= 3(2) + 3(6) + 3(12) + 3(20) + 3(30) + 3(42)

= 6 + 18 + 36 + 60 + 90 + 126

= 336

6

Therefore, ∑ 3(n + n2) = 336

i=1

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Example 3: Solve and find the sum for the given expression;

6

∑ n(2n + 1) / 2 =?

n=1

Solution:

6

∑ n(2n + 1) / 2 =?

n=1

Formula: n(2n + 1) / 2, starting term: 1, end term: 6, and index variable: n

6

∑ n(2n + 1) / 2 = (1(2(1) + 1) / 2) + (2(2(2) + 1) / 2) + (3(2(3) + 1) / 2) + (4(2(4) + 1) / 2)

n=1  + (5(2(5) + 1) / 2) + (6(2(6) + 1) / 2)


= (1(3) / 2) + (2(5) / 2) + (3(7) / 2) + (4(9) / 2) + (5(11) / 2) + (6(13) / 2)

= (3 / 2) + (10 / 2) + (21 / 2) + (36 / 2) + (55 / 2) + (78 / 2)

= 1.5 + 5 + 10.5 + 18 + 27.5 + 39

= 101.5

6

Therefore, ∑ n(2n + 1) / 2 = 101.5

n=1
Solving Summation Notation – Solving Practice Problems

Problem 1: Solve and calculate the sum for the given expression;

5

∑ n2(n) =?

n=1

Answer: 225

Problem 2: Solve and calculate the sum for the given expression;

5

∑ (2n + 2) / n =?

n=1

Answer: 14.57

Define Expanded Form

Expanded Form is a method to break up a number to illustrate how much each digit in the number represents and in other words, expanded form is the method of pulling a number separately and expressing it as a sum of the values of every digit. In this article we study about expand form define of and develop the knowldge of the math .

Example on Define Expanded Form

Choose the expanded form of the number 15,793.

Solution:

The expanded form of the number 15,793 can be found using a place value chart.

Now, let’s aim to write the number in words and define how it helps us identify the expanded form.

The number 15,793 is read as: 15 thousands 7 hundreds 93 which is the same as 1 ten thousand  5 thousands 7 hundreds 9 tens and 3 ones.

So, the expanded form of 15,793 is 10,000 + 5,000 + 700 + 90 + 3.

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Example to Define Expand Form Math:

Example 1:

57  can be expand as 57 = 50 + 7.

226 can be expand as 296 = 200 + 20 + 6.

5467 can be expand as 5467 = 5000 + 400 + 60+7.

89192 can be expand as 89192 = 80000 + 9000 + 100+90+2

Example 2:

What will we get when we expand form 23,368?

Choices:
A. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones
B. 2 ten thousands, 3 thousands, 6 hundreds, 6 tens, and 7 ones
C. 5 thousands, 5 ten thousands, 6 hundreds, 6 tens, and 8 ones
D. 5 thousands, 8 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 23,368 = 20,000 + 3,000 + 300 + 60 + 8
Step 2: = (2 × 10,000) + (3 × 1,000) + (3 × 100) + (6 × 10) + (8 × 1)
Step 3: So, when we expand 23,368, we will get ‘2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones’.

Example 3:

What will we get when we expand form 13,369?

Choices:
A. 1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones
B. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 5 ones
C. 3 thousands, 7 ten thousands, 9 hundreds, 6 tens, and 9 ones
D. 4 thousands, 7 hundreds, 9 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 13,369 = 10,000 + 3,000 + 300 + 60 + 9
Step 2: = (1 × 10,000) + (3× 1,000) + (3 × 100) + (6 × 10) + (9 × 1)
Step 3: So, when we expand 13,368, we will get ‘1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones’.

Systematic Sampling Method

In this article, let us study what is sampling and the types of sampling.

Preliminary concepts of sampling:

In testing the quality of bulbs produced, by a company it is impossible to test every bub manufactured by a company; it is quite sufficient if a sample is taken for testing and based on this test it is possible to draw conclusion about the population. If each and every item is tested, in some cases it may not be possible to send them to the market. thus, process of sampling is to get information about the population from the sample. To the extent to which we can do this with any accuracy depends on the choice of our sample or samples.

Samples are classified as follows:

non-probability samples

probability samples

Some non -probability samples  are 1. deliberate sampling , 2. Quota sampling 3. Block sampling.

Some probability samples  are1. Random sampling , 2. Stratified sampling , 3. Systematic sampling , 4. Multi stage sampling
What is Systematic Sampling

Systematic sampling with a random beginning is a form of restricted random sampling which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in a serial order and every ith element, starting from any of the first i items is chosen.

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Example for Systematic Sampling:

Suppose we require 5% sample of students from a college where there are 2000 students and where everyone is given a departmental number from 1 to 2000, we should first select one number at random from 1 to 20. Suppose the number chosen is 12; then our sample consists of students with departmental numbers 12, 32, 52, 72,..1992

If, however, the students are arranged in groups of 20 such that the first student belongs to the first of 20 specified communities, the second to a second community and so on, then our sample may consist of students of all belonging to the same community and consequently we cannot consider the sample to be random

Integral of ln u

Integral calculus is a branch of calculus that deals with integration. If f is a function of real variable x, then the definite integral is given by

int_a^bf(x)dx

Where a and b are intervals of a real line.

Integral of ln u means integration of natural logarithmic (ln) function with respect to the variable ' u '. In this article, we are going to see the list of natural logarithmic integral (ln) rules with few example problems.
List of Natural Logarithmic (ln) Integral Rules:

int 1/u dx = ln |u| + C

int  ln u du = uln u  - u + C

int uln u du = (u^2)/2 ln u  - 1/4 u2 + C

int u2 ln u du = (u^3)/3 ln u - 1/9 u3 + C

int (ln u)2 du = u(ln u)2 - 2uln u + 2u + C

int ((ln u)^n)/u dx = 1/(n + 1) (ln u)n+1 + C

int  (du)/(uln u) = ln (ln u) + C

Learning Natural Logarithmic (ln) Integral Rules with Example Problems:

Example problem 1:

Integrate the function f(u) = 5/u

Solution:

Step 1: Given function

f(u) = 5/u

int f(u) du = int5/u du

Step 2: Integrate the given function f(u) = 5/u with respect to ' u',

int5/u du = 5ln (u)+ C

Example problem 2:

Integrate the function f(u) = log (9u)

Solution:

Step 1: Given function

f(u) = log (9u)

int f(u) du = intlog (9u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let x = log (9u)    dv = du

dx = 1/u  du         v = u

int x dv = xv - int v dx

int log (9u) du = ulog (9u) - int u 1/u du

= ulog (9u) - int du

= ulog(9u) - u + C

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Example problem 3:

Integrate the function f(u) = 10log (3u)

Solution:

Step 1: Given function

f(u) = 10log (3u)

int f(u) du = int10log (3u) du

The above function can be written as

int f(u) du = 10intlog (3u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let u = log (3u)    dv = du

du = 1/u  du         v = u

int x dv = xv - int v dx

10 int log (3u) du = 10[ulog (3u) - int u 1/u du]

= 10ulog (3u) - 10int du

= 10ulog(3u) - 10u + C

Equilateral Pentagon Angles

In geometry an equilateral pentagon is a polygon with five sides of equal length. Its five internal angles, in turn, can take several values, thus permitting to form a family of pentagons. In contrast, the regular pentagon is unique, because is equilateral but at the same time its five angles are equal. Four intersecting equal circles disposed in a closed chain, are sufficient to describe an equilateral pentagon. ( Source Wikipedia )


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Examples

Problems 1

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 100, 58 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=100

Angle 3 a3=58

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+100+58+67+x=540

250+125+x=540

375+x=540

Subtract 375 on both sides

375-375+x=540-375

x=165

The missing angle is 165

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Problems 2

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 70, 158 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=70

Angle 3 a3=158

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+70+158+67+x=540

220+225+x=540

445+x=540

Subtract 375 on both sides

455-455+x=540-455

x=95

The missing angle is 95

Problems 3

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 100, 150, 90 and 40

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=100

Angle 2 a2=150

Angle 3 a3=90

Angle 4 a4=40

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

100+150+90+40+x=540

250+130+x=540

380+x=540

Subtract 375 on both sides

380-380+x=540-380

x=160

The missing angle is 160

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Problems 4

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 155, 120, 60 and 70

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=155

Angle 2 a2=120

Angle 3 a3=60

Angle 4 a4 =70

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

155+120+60+70+x=540

275+130+x=540

405+x=540

Subtract 375 on both sides

405-405+x=540-405

x=135

The missing angle is 135

Distance Formula Proof

For distance formula proof statistics will be deals with the co-ordinates parameters.Formula for distance calculation by the co ordinates of the points given. For calculating distance statistics between the points we have the distance formula proof and we substitute the points in that formula we find the distance between the points.In this article we have the distance formula proof and the problems using the distance formula.
Distance Formula Proof:
Distance between two points co ordinates is a basic concept in geometry.Now, we give an algebraic expression for the same.

Let P1  (x1, y1) and P2 (x2, y2) be two points in a Cartesian plane and denotes the distance between P1 and P2 by d(P1, P2) or  by  P1P2. Draw the line segment ` bar(P_1P_2)`

distance formula proof

The segment ` bar(P_1P_2)` is parallel to the x axis  Then y1 = y2. Draw P1 L and P2 M, perpendicular to the x-axis. Then d(P1,P2) is equal to the distance between L and M. But L is (x1, 0) and M is (x2, 0).

So the length LM = |x1-x2| Hence d (P1, P2) = |x1-x2|.

therefore, [d(P1,P2)]2= |x1-x2|2+ |y1-y2|2

=(x1-x2)2+(y1-y2)2

=(x2-x1)2+(y2-y1)2
d(P1,P2) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Problems Using Distance Formula Proof:
Example 1:
Find the distance between the points A(-2,3) and B(3,7).

Solution:
Assume d be the distance between A and B.             (x1,y1) = (-2,3)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                   (x2,y2) =  (3,7).

=`sqrt((3+2)^2 +(7-3)^2)`

=`sqrt(5^2+4^2)`

=`sqrt(25+16)`

=`sqrt41`
Example 2:
Find the distance between the points A(-1,1) and B (3,3).

Solution:
Assume d be the distance between A and B.                (x1,y1)= (-1,1)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                                (x2-y2)= (3,3).

=`sqrt((3+1)^2 +(3-1)^2)`

=`sqrt(4^2+(2)^2)`

=`sqrt(16+4)`

=`sqrt20`
=2`sqrt 5`

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Example 3:
Find the co ordinates distance between the points A(-1,5) and B (1, 4).

Solution:
Assume d be the distance between A and B.                       (x1,y1)= (-1,5)

Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                          (x2-y2)= (1,4).

=`sqrt((1+1)^2 +(4-5)^2)`

= `sqrt(2^2+(-1)^2)`

=`sqrt(4+1)`
=`sqrt5`