# Basic Probability Concepts Assignment help

## Basic Concepts of Probability

**Introduction of solving basic probability**

Basic probability is a way of expressing information or belief that an event will occur or has occurred. In mathematics the concept has been given a correct meaning in basic probability theory that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems.

Formula used for solving basic probability

= Number of favourable out comes

Total Number of favourable outcomes

**Types of Probability in Basic Concepts of Probability and Statistics**:

The probability divided into three types. The basic concepts types are as follows,

- Classical theory
- Frequency of occurrence
- Axiomatic probability theory

**Basic Measurement Levels in Basic Concepts of Probability and Statistics:**

- Nominal
- Ordinal
- Interval
- Ratio

## Solving Problems Based on Basic Probability:

We will use the above formula for solving basic probability.

Pro 1:In basic probability, what are the odds in favours of getting a ‘4’ in throw of a die? What are the Events against getting a ‘4’?

Solution:There is only one outcome favourable to the event “getting” a 3, the other five outcomes, namely, 1, 2, 3, 4, 5, 6 are unfavourable in basic probability. Thus,

Events in favour of getting a ‘4’

= Number of favourable out comes

Number of outcomes

=`1/6`

Events against getting a ‘4’

= Number of unfavourable out comes

Number of outcomes

=`5/6` .

Pro 2: If the odds in favour of an event are 3 to 5, find the basic probability that it will occur.

Solution:The odds in favour of the event are `3/5`3/5. Thus,

P(A)= 1-P(A') = 1 - `3/5`

P(A) = `2/5`

The basic probability that it will occur = `2/5`

Pro 3:Sam and Alex appear for an interview for two vacancies. The basic probability of Arum’s selection is ¼ and that of Alex’s selection is 1/6. Find the basic probability that

i) Only one of them will be selected,

ii) None of them be selected.

Solution:Let A:Sam is a selected B: Alex is selected. Then, P(A)=`1/4` and P(B)=`1/6`

Clearly ‘A’ and ‘not B’ are independent also ‘not A’ are independent.

i) P(only one of them will be selected)

=P(A are not B or B and not A)

=P(A) P(not B) +P(B)P(not A)

=`1/4 * ( 1 - 1/6) + 1/6 ( 1- 1/4)`

=`1/4 * 5/6 + 1/6 * 3/4`

=`5/24 + 3/24`

=`8/24 = 1/3`

ii) P(only one of them be selected)

=P(not A and not B)

=P(not A)xP(not B)

=`(1 - 1/4) * ( 1- 1/6)`

=`3/4 * 5/6 = 15/24 = 5/8`

## Solving Practices Problem for Basic Probability:

Pro 1:In a simultaneous toss of two coins, find the basic probability of 2 heads.

Ans: `1/4`

Pro 2:What are the odds in favours of getting a ‘2’ in throw of a die? What are the Events against getting a ‘2’?

Ans: `1/6` and `5/6`

## More Examples

**Example 1:**

** **There are 100 boys are in the school. In those boys, there are 30 boys are playing cricket, 20 boys are playing tennis and the remaining boys are playing football. Find the probability for the following conditions?

i) Select the boys who playing the cricket.

ii) Select the boys who playing the football.

**Solution**

Total number of boys n(S) =100

Number of boys playing the cricket n(A)= 30

Number of boys playing tennis n(B)= 20

Number of boys playing football n(C) = 100 - (30 + 20)

= 100 - 50 = 50.

i)Assume P(A) is the probability for select the students who playing the cricket.

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

= `30/100`

= `3/10` .

ii) Assume P(C) is the probability for select the boys who playing the football.

P(C) = `(n(C))/(n(S))`

P(C) = `50/100`

=`1/2`

## Simple Probability

**Introduction simple probability:**

In mathematics simple Probability is one of the subdivision in that the chance of a result happening is assign to a mathematical value which predicts that result is to occur.

In simple probability there is **'n' ** simple events mingle by means of an approximate experiment and **'m' **of them are approving to an event A, then the probability of result coming is

** 'A' ** is distinct as the ratio `m/n` and is denote by **P (A).**

** ** Hence** P (A) = `m/n` ****.**

## Simple Probability Fundamental Concepts and Formulae

1.) Probability is a simple quantitative quantify of assurance.

2.)** Random experiment:** Any movement which is linked to certain outcome is called approximate experiment, e.g.

(i) Tossing a coin (ii) Throwing a die.

3.) **Elementary events:** A result of an approximate experiment is called an elementary event or basic event.

4.) **Sure events:** Those events which probability is one.

5.) **Impossible Events:** Those events which probability is 0.

6.) Probability of several events forever fabrication between 0 and 1, that is

i.)Probability of an occurrence can’t be negative.

ii.)Probability of an occurrence can’t be except one.

7**.) Negation of event**: Related with each occurrence a linked with an approximate experiment we classify an event “**not A**” or (**A** bar) that occur only when A doesn’t occur.

8.) For a few events A, P (A) + P(**A** bar) = 1 = P(A) = 1 – p(A).

9.)** Compound event**: An occurrence linked to an approximate experiment is a compound event if it is achieve by two or more basic events linked to the approximate experiment.

10.) **Occurrence of an event**: A result linked to an approximate experiment is said to happen if any of the basic events linked to the event A is an outcome.

## Simple Probability Example Problem:

In a simple probability a game consists of tossing a coin three times and note its outcome every time. Raghul wins if every tosses give the equal result i.e., 3 heads or 3 tails, and loses or else. Evaluate the probability of Raghul will lose the game.

**Solution:**

The outcomes linked with this trial is given by

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

** Hence** Total number of likely outcomes = 8

Now, Raghul lose the game if he gets

HHT, HTH, THH, TTH, THT, HTT

** Hence** positive number of events = 6

** Hence** probability that he lose the game = 6 / 8 = 3 / 4

## Basic Concepts:

The math practice 4 involves some basic concepts in mathematics. In this we are going to see about algebra in math practice 4 .They are:

- Variables
- Expressions
- Terms
- Polynomials

- Equations

## Example Problems:

**Example 1:** Solve for x?

79 + 33 = x

**Solution:**

To find x value

79 + 33 = x

x =112

The answer x is 112

**Example 2:**

The number decreased by 70 is 12 times its opposite. Find the number.

**Solution: **First we convert the algebra word problem to a numerical form to get the solution.

The number is decreased by 70 is 12 times its opposite.

Write an equation.

x - 70 = -12x (Equation)

x - 70 + 70 = -12x + 70 (add 70 on both sides we get)

x + 12x = -12x +12x + 70 (add 12x on both sides we get)

11x = 70

x = 6.36

**Example 3 :** Solve for M?

77 + 3 = M

**Solution:**

77 + 3 = M

M = 80

The answer M is 80

**Example 4:** Solve for q?

q + 7 = 49

**Solution:**

To solve the p value

q = 49 – 7

q = 42

**Example 5:** Solve the following system of the linear equations using the method of substitution.

x - y = -6 ,4x+8y = -48

**Solution:**

**Step 1: **Rearrange the first equation,

x - y = -6

y = x + 6

** Step 2:** Substitute this value for y into the second equation;

4x + 8(x + 6) = -48

** Step 3:** Expand and simplify the equation:

4x + 8x + 48 = -48

12x = -96

x = -8

** Step 4:** Substitute x back into the one of that original equations;

-8 - y = -6

y = -2

**Submit Assignment**

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