Solving Probability Practice

Introduction to Probability:
Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.For solving probability practice problems we must know the following formula,

Probability of an event   =    number of times an event occurs / total number of outcomes

Consider an observation with 'n' possible ways and out of them in 'm' ways if the event 'A' occurs,then the probability of occurrence of the event 'A' is given by P(A) = m/n.

Let us workout some practice problems on solving probability in the following sections.

I like to share this Probability Problems and Solutions with you all through my article.

Probability on a coin problem:


Here Tossing a coin is a random experiment. When you toss a coin, you may get head or tail.Let us see how to  solving the coin problem.

1: What is the probability of getting Head when you toss a coin?

We can have 2 out-comes for tossing a coin :  Head or tail.

Here the event is getting Head. So we have 1 head.

So probability of getting head is one out of two outcomes.

Probability of getting Head    = number of head event occurs / total number of outcomes

= 1/2

= 0.5

So the probability of this practice problem is 0.5


Probability on Card problem:


In 52 cards, there are 13 spades, 13 clovers, 13 diamonds and 13 hearts , 26 cards are red in color (diamonds and heart) and 26 are black (spade and clover).Let us see how to  solving the card problems.

(1)  What is the probability that you get a red card?

The event is getting a red card. There are 26 red cards.

Total sample spaces are 52(52 cards).

Probability of getting a red cards = number of red cards / sample space

= 26/52

=1/2

= 0.5

(2) What is the probability of getting a spade when you draw from a well shuffled pack of 52 cards?

Number of Spade cards = 13

Total Sample space       = 52

Probability of getting a spade card = number of spade / total sample space

= 13/52

= 1/4

= 0.25

(3) What is the probability of getting a black queen in a pack of 52 cards?

In a pack of 52 cards, there are 2 Black Queens (spade 1, clover 1)

Total Sample space = 52

Probability of getting a black queen = number of black queen / total sample space

= 2/52

= 1/26

= 0.04

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Practice Problem:


The following practice problems on solving probability will helps for good understanding.

Practice problem 1: What is the probability of getting Tail when you toss a coin?

Practice problem 2: What is the probability of getting a diamond when you draw from a well shuffled pack of

52 cards ?

Answer key:

Practice problem 1:  0.5

Practice problem 2: 0.25

Study Practice Algebra Problems

Introduction of Study Practice Algebra Problems:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, algebra is one of the main branches of pure mathematics. (Source: Wikipedia)

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Study Example Algebra Problems:


Study practice algebra problem 1:

Simplify the equation 5(a -3) + 4b - 7(a -b -3) + 5.

Solution:

5(a -3) + 4b - 7(a -b -3) + 5

Multiply factors.

5a - 15 + 4b -7a + 7b + 21 + 5

Group like terms.

-2a + 11b + 11

Answer:   - 2a + 11b + 11

Study practice algebra problem 2:

To find the 4x - 8y = 9 equation on x-intercept.

Given:

4x - 8y = 9

Solution:

4x - 8y = 9

To find the x intercept of y = 0 and solve for x.

4x - 0 = 9

Solve the value of x.

x = 9 / 4

The x intercept is at the point (9/4, 0).

Answer: x intercept is at the point (9/4, 0).

Study practice algebra problem 3:

16x + 2y = 32.

y + 8 = 16x

Find the x and y value.

Solution:

16x + 2y = 32 (equation 1)

y + 8 = 16x   (equation 2)

Step 1: To choose the equation where the coefficient of the variable is 1.Choose equation 2 to isolate the variable y

y = 16x – 8 (equation 3)

Step 2: From equation 3, we know that y-value is the same as the 16x – 8. We can to substitute the variable y-values in the equation 1 with 16x – 8.

16x + 2 (16x – 8) = 32

Step 3: Remove the brackets by using the distributive property

16x + 32x – 16 = 32

Step 4: To combine the terms

48x – 16 = 32

Step 5: To Isolate the variable of x

48x = 48

x = 48 / 48

x = 1

Step 6: Substitute x = 1 in the equation 3 to get the y-value.

y = 16 (1) – 8

=16 – 8

y = 16 - 8

y = 8

Step 7: To check the answer with the equation (1)

16(1) + 2 (8) = 16+ 16 = 32

Answer: x = 1 and y = 8

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Study Practice Algebra Problems:


Practice Problem 1:

Find the x and y value 12x + 2y = 20, y + 8 = 12x

[Answer: x = 1 and y = 4]

Practice problem 2:

Solve for x and y for the following equations 5x + y = 8, 6x + y = 7

[Answer: x = -1 and y = 13]

Practice Problem 3:

Simplify the equation 7(x -3) + 5y - 3(x -y -3) + 5.

[Answer:   4x + 8y – 7]

Practice Algebra Class

Introduction to practice algebra class:

Algebra class deals with the Pre-algebra, Functions and Graphs, Integers, Rational Numbers, Equations in One Variable, Equations in Two Variables, Simultaneous Equations, Problem Solving etc. in arithmetic, we use digit like 1,2,3, etc to represent numbers. In algebra we use numbers as well as letters of the alphabet such as a, b, c, etc for any numerical values we choose. The basic thought of the algebra is balancing the algebraic equations on both sides of the give equations in the problems. Algebra will perform the four fundamental operations such as addition, subtraction, multiplication and division. In this article we shall discuss with practice algebra class


Sample problem on practice algebra class


Problems of algebra class on linear equations:

Example 1:

Solve the linear equation for algebra class

11 x - 15 = 19 x - 23

Solutions:

11 x - 15 = 19 x - 23

Subtract 19 x from both sides of the above equation, we get the below term

-8x – 15 = -23
Add 15 to both sides of the above equation, we get the below term

-8x – 15 + 15 = - 23 + 15

-8x = - 8
Divide both sides by -8 of the equation, we get

X = 1

The answer is x = 1

Test the solution for the given equation. Replace with in the above given equation for x. If the left side of the equation sum value is equal to the right side of the equation sum value after the replacement of x value, you have got the exact answer.

Left side of the equation:
11(1) - 15 = - 4

Right side of the equation:
19(1) - 23 = - 4

Practice word problems of algebra class

Example 2:

A bag contains ten, five and two rupees currencies. The total number of Currencies is 30 and the total value of money is Rs.135. If the second and third currencies are interchanged the value will decrease Rs.8. Find the number of currency in each sort.

Solution :

Let x, y, z be the number of Rs.10, Rs.5 and Rs.2 currencies respectively.

The total number of currencies is 30 => x + y + z = 30--------(1)

If the total value of money is Rs.135 => 10x + 5y + 2z = 136--------(2)

If the II and III of currencies are interchange the value will be decreased by Rs.8

=>10x + 2y + 5z = 135 – 8 = 127----------------(3)

10x + 2y + 5z = 127

10 x (1) – (2) =>     5y + 8z = 106------------(4)

(2) – (3)        =>     3y - 3z = 9

y – z = 3------------------(5)

Let us solve (4) and (5)

(4) =>      5y + 8z = 106

8 x (5)   ⇒ 8y – 8z = 24

13y = 130

y = 10

Substituting y = 10 in (5) we get

z = 7

Substituting z = 7, y = 10 in (1) we get

x + 10 + 7 = 30 => 30 – 10 – 7 or x = 13

Number of Rs.10, Rs.5 and Rs.2 currencies are respectively 13, 10 and 7.

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Practice problem for algebra class:

Problem 1:

2x + 11 = 20 - x

Answer: x = 3

Problem 2:

7x + 11 = 19 + 3x


Answer: x = 2

Statistics Practice Exam

Introduction to statistics practice exam:

Statistics is the science of making effective use of numerical data relating to groups of individuals or experiments. It deals with all aspects of this, including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments.

A statistician is someone who is particularly versed in the ways of thinking necessary for the successful application of statistical analysis. Often such people have gained this experience after starting work in any of a number of fields. There is also a discipline called mathematical statistics, which is concerned with the theoretical basis of the subject.

Let us see several statistics based solved problems and statistics practice exam. Statistics practice exam with solutions which would be helpful for students to evaluate their performance.

Concepts on statistics practice exam:


Before going for a statistics practice exam, let us recall how to find mean deviation. The steps to find the mean deviation are as follows.

Step 1: find the mean for given data
`barx` = (sum of all variables) /( given number of  variables).

Step 2: find the deviation of the respective observations from mean `barx` ,  in other words  xi -  `barx`
Step 3: find the modulation of deviation ,     | Xi - `barx` |
Step 4: mean deviation of the mean,
M.D.(`barx` ) = `sum` n i-1 | xi - x| / n

Example1:

Find the mean deviation about the mean for the following data:

12, 5, 16, 14, 11, 5, 10, 15

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(12+5+16+14+11+5+10+15)/8` = `88/8` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

12–11,5–11,16–11,14–11,11–11,5–11,10–11,15–11,

or 1,–6,5,3,0,–6,–1,4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

1,6,5,3,0,6,1,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(1+6+5+3+0+6+1+4)/8` = `26/8` = 3.25.

Example2:

Find the mean deviation about the mean for the following data:

15, 5, 12, 16, 12, 4, 9, 16,7.

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(15+5+12+16+12+7+9+16+7)/9` = `99/9` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

15–11,5–11,12–11,16–11,12–11,7–11,9–11,16–11,7-11.

or 4,–6,1,5,1,–4,–2,5,-4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

4,6,1,5,1,4,2,5,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(4+6+1+5+1+4+2+5+4)/8` = `32/8` = 4.

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Practice on statistics practice exam:

Statistics Practice exam:

1. Find the mean deviation for the mean of the following data:
6, 7, 10, 12, 13, 4, 8, 12

2. Find the mean deviation for the mean of the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5

Answers for the above questions of statistics practice exam:

1.   2 75

2.   6.2

Practice Abstract Algebra Exams

Abstract for algebra:

In algebra the algebraic expressions represent mathematical ideas and operations in a short way using symbols and variables.The algebra equation states that one expression names the same number as another expression. The symbol in algebra which can be assigned different numerical values is called a variables. Algebra uses known quantities to find the unknown quantities. Here it is about the practice for abstract algebra exams. I like to share this algebra 2 problems and solutions with you all through my article.


practice abstract algebra exams - Examples:


practice abstract algebra exams - Problem 1: Simple equations:

solving simple equation: 7x + 13 = 4x + 43.

Solution:

since 7x is larger than 4x,

we decide to shift the letter terms to left and the number terms to right side.

7x + 13 = 4x + 43

we want to remove 4x from the right side,

we can do so by subtracting 4x from both sides.

7x + 13 – 4x = 4x + 43 – 4x

3x + 13 = 43

we want to remove 13 from both sides,

we do so by subtracting 13 from both sides.

3x + 13 – 13= 43 – 13

3x = 30

Dividing both sides by 3.

3x/3 = 30 / 3

x = 10

check to substituting the x in as:

left side: 7(10) + 13 = 70 + 13 = 83

right side: 4 (10) + 43 = 40 + 43 = 83

practice abstract algebra exams - Problem 2: Algebraic expressions:

Simplify the algebraic expression: 17m² – 9m + 5m – 3m² – 7m + 11

solution:

By rearranging the given terms as,

we have,

=17m² – 3m² + 5m – 9m – 7m + 11

By equating, we get

= (17 – 3) m² + (5 – 9 – 7) m + 11

To simplify the equation as

=14m² + (– 4 – 7) m + 11

=14m² + (–11) m + 11

=14m² – 11m + 11.

This is the equation for the given algebraic expression.

practice abstract algebra exams - Problem 3: Algebraic identities:

Evaluate 102 × 104 without directly multiplying the two given numbers.

Solution:

Taking 102 as (100 + 2) and 104 as (100 + 4)

we get the multiplier as

= 102 × 104

Taking the multiplier as

= (100 + 2) × (100 + 4)

By multiplying and adding the identities we get,

= 100² + (2 + 4) 100 + 2 × 4

= 10000 + 6 × 100 + 8

= 10000 + 600 + 8

= 10608

This is the solution for the given algebraic identities.

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practice abstract algebra exams - Problems for practice exam:


1. solving simple equation:18 – 5x = 3x – 6.

Answer: x = 3

2. Evaluate the algebraic identities 103 × 105 without directly multiplying the two given numbers.

Answer: = 10815

3. Simplify the algebraic expression: 16m² – 9m + 5m – 3m² – 7m + 12.

Answer: =13m² – 11m + 12.

Study Maths Advantage

Introduction to study maths advantage:

The study of maths is important nowadays in everyday life.  For developing the skills of the individual persons, maths is very important. Every industry needs mathematics today. Generally math is invented for the purpose of real time application for calculation purpose. In this article,  we are going to see about the maths advantage using real time application with some example problems.

Advantage to study maths

Advantage:

Employment
Science
Technology
Medicine
Example for the study of maths advantage:

In computing industry, complex programs are created using the mathematics.
In Cryptography, the encode and decode are done using the mathematics.
In Internet, credit and debit card are also done using the maths.
For calculating the area of land, we have formula in mathematics. By using this formula, it will be easy to calculate the area.
Speed can also be calculated using the formula in mathematics.
Distance can also be calculated using the formula in mathematics.

Example problems for study maths advantage:

Real Time Example:

Problem 1: Find the area of the given square, where, width = 24, height = 35.

Solution:

Step 1:Given:

Width = 24, Height = 35

Step 2: To Find:

Area of the square.

Step 3: Formula:

Area of the square = width `xx` Height

Step 4: Solve:

Area of the square = width `xx` Height

= 24  `xx` 35

= 840

Result: Area of the square = 840

Problem 2: Find the speed of the given car, where, distance = 345, Time = 30.

Solution:

Step 1:Given:

Distance = 345

Time = 30.

Step 2: To Find:

Speed of the car

Step 3: Formula:

Speed of the car = `(Distance)/(Time)`

Step 4: Solve:

Speed of the car = `(Distance)/(Time)`

= `345/30`

= 11.5

Result: Distance of the car = 11.5

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Practice problems for study maths advantage:


Problem 1: Find the speed of the given car, where, distance = 240, Time = 20.

Answer: 12

Problem 2:  Find the area of the given square, where, width = 34, height = 46.

Answer: 1564

Pre Algebra Practice Tests

Introduction to online pre algebra practice tests :

Algebra , most important topic in mathematics and it’s very easy to calculating, concerning the study of structure, quantity and relation. We can use arithmetic operation and the letter and symbols are referred to as Variables.  Algebra is derived from Arabic language Al-jabr. Algebra also performs addition, subtraction, multiplication, division. For example,   2+3=3+2 is same as p+q=q+p. Let us see about online pre algebra practice tests.


Problems for pre algebra:


Example 1 for online pre algebra practice tests problem:

Multiply -2a2bc, 2a2b and −1/4

Solution for online pre algebra practice tests problem:

=>( - 2a2bc ) ( 2a2b ) ( -1/4 )

=>( -4 a4b2c ) ( - 1/4 )

=>a4b2c = a2bc * a2b

Example 2 for online pre algebra practice tests problem:

Find factors of 3x2y

Solution for online pre algebra practice tests problem:

=>( 1 * 3x2y ) (3x2y ) ( 3x * xy )  (3xy * x ) ( x2 * 3y ) (y * 3x2 )

=>(1,3x2y), (3x2y), (3x,xy), (3xy, x),(x2 ,3y),(y,3x2 ).

Example 3 for online pre algebra practice tests problem:

Simplify : ( 2x + 2) ( 4x – 10 ) +  25

Solution for online pre algebra practice tests problem:

First of all remove the paranthesis by multiplying the terms inside the bracket.

=>8x2 – 20x + 8x – 20 + 25

=>8x2 – 12x + 5

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More Example Problems


More Example Problems For Online Pre Algebra Practice Tests:

1.Evaluate (a) (2x + 3y)2 (b) (3 / 2x − 6y)2 (c) (x2 + 2y2 )(x2 − 2y2 )

Answer: (a)4x2 +12xy + 9y2 (b) 9/4x2 18x y 36y2 (c)x4 − 4y4

2.Factorize 6x3 + 8 x2y.

Answer:2x2×3x+2x2×4y = 2x2(3x+4y)

3.Find the greatest common factor of 2x5 and 12x2 and then factorize 2x5+12x2

Answer:2x2 (x3 + 6)

4.Factorize 4m2 - 4m +1.

Answer:(2m - 1)2 =(2m- 1) (2m-1)

5.Factorize x2- 2xy + y2 – 9

Answer:(x-y + 3) (x-y-3)