MATHEMATICS IN EASY STEPS : PROBABILITY AND ITS APPLICATIONS ( WITH SIMPLE EXAMPLES FOR UNDERSTANDING )

  "Mathematics is a easy subject but maximum find them so hard to learn and remember Mathematics functions and rules. That all because we not give time to understand basics of Mathematics rule. So learn here the basic rules step by step."

PROBABILITY

"Probability is a numerical description of possibilities that an event can be occur."

The probability of an event is a number between 0 and 1. '0' indicates impossibility of the event and '1' indicates certainty to find the probability of a single event to occur we need to know the total number of possible outcomes. 

Probability can be written as decimal value or in percentage.

The possibility of an event will be written as P ( E ) .

1. If P ( A ) › P (B) ; then event A has higher chance of occurring then event B.

    If P ( A ) = P (B) ; then event A and B are equally likely to occur.

2. If an event may happen 'X' ways and fails to happen in 'Y' ways in a trial , 

    Then the probability to event happening P ( E ) = X∕(X+Y) , and

   The probability to event fails to happening = 1- P (E)

                                                                     = 1- ⟨ X∕(X+Y) ⟩

FORMULAS :

1. Probability of an event occur P ( E ) = ( Number of Favorable Outcomes )/                                                                            ( Total Number of Outcomes)

2. Probability that the event will not occur P ( E' ) = 1- P ( E )

3. P ( E ) + P ( E' ) = 1   

       

SOME EXAMPLES FOR UNDERSTANDING :   

Example 1.

    If we tossing a coin then :-

(i). Probability to find Head P (H) = 1/2 = 0.5 = 50 %

(!!). Probability to find Tail P (T) = 1/2 = 0.5 = 50 %  

    Here,  P (H) = P (T)  

Example 2 .

    If coin tossing 2 times then, 

Total possible events are HH, TT, HT, TH = 4 events

       Hence, 

(i). Possibilities to find 2 Head P (H₂) = 1/4

(ii). Possibilities to find 2 Tail P (T₂) = 1/4

(iii). Possibilities to find one Head and one Tail P (HT) = 2/4 = 1/2

Example 3 .

    A dice rolled one time then total possible outcomes will be 1,2,3,4,5,6 = 6

    Hence,

(i). Possibilities to find 1 = P (1) = 1/6

(ii). Possibilities to find 1or 5 = P (1or 5) = 2/6 = 1/3

(iii). Possibilities to find 1or 3 or 5 = P ( 1or 3 or 5 ) = P ( ODD )= 3/6 = 1/2

(iv). Possibilities to find 2or 4 or 6 = P ( 2or 4 or 6 ) = P ( EVEN )= 3/6 = 1/2

Example 4 

    There are 6 books in a bag. 2 are English, 3 are Maths, 1 is science. What is possibility to get an English book.

Solution :

    Here , No. of favorable chances to get English book = 2

               Total number of outcomes = 2+3+1 = 6

    Hence,

                P ( English ) = 2/6 = 1/3

Example 5 .

    Two coins are tossed 300 times and find -

        2 Heads = 100 Times,

        1Head = 50 Times,

        No Head = 150 Times.

Then;

    Probability to get 2 Heads = P (E₁) = 100/300 = 1/3

    Probability to get 1 Head = P (E2) = 50/300 = 1/6

    Probability to get No Head = P (E3) = 150/300 = 1/2

Example 6 .

    The record of a weather station shows that out of past one year , its weather forecast were correct 225 times. Then--

(!). The probability that on a given day it will be correct  P (E) 

        = (No. of days when the forecast was correct) / (Total no. of days of forecast)

        = 225 / 365  = 0.62 = 62%

(ii). The probability when the forecast will not br correct P (E') 

        = 1 - P (E)

        = 1-0.62 = 0.38 = 38%

                            OR

        = (365 - 225) days / 365 days  = 140 / 365  = 0.38 = 38%

NOTE :  Always P (E) + P (E') = 1 or 100%

LAWS OF PROBABILITY

There are three important Laws of Probability, they are under below :--

1. Addition Law of Probability

2. Multiplication Law of Probability

3. Law of Conditional Probability

1. Addition Law of Probability :

   

 " The probability of any event is the sum of the probability of its mutually exclusive events (sub events). "

   If the two events "A" and "B" are mutually exclusive and,

    P (A) = Probability of occurrence of event A ,

    P (B) = Probability of occurrence of event B.

   Then the probability of occurrence of the compound (Total or whole) event A+B is -

        

        P ( A+B ) = P (A)+P (B)

That we can also explain following way :-

        P ( A∪B ) = P(A)+ P(B)-P ( A∩B )

Here A and B are disjoint , Hence P ( A∩B ) = 0

So that,

            P ( A∪B ) = P(A)+ P(B)     (For exclusive events)

EXAMPLE :

    In a bag contains 200 balls, in which 80 red balls, 50 white balls, 70 blue balls.

If we draw one ball randomly from bag. Then-

(i). The probability that will be red (P_R ) = 80/200 = 0.4

(ii). The probability that will be white (P_W ) = 50/200 = 0.25

(iii). The probability that will be blue (PB ) = 70/200 = 0.35

(iv). So, that the probability that it will be red or white or blue

                = PR + P_{W +} PB   =  0.4 + 0.25 + 0.35  =  1

2. Multiplication Law of Probability :

     " If a compound event be made up of number of separate and independent events, and occurrence of the compound event be the result of each of these sub-events happening, the probability of occurrence of the compound event is the product of the probabilities that each of these sub events will happen."

So, where it is possible for two or more sub events to occur together, the probability they will do so, is the product of their individual probability.

        P ( A . B ) = P(A) + P(B)

3. Law of Conditional Probability :

      When the events are not independent but the nature of the dependence is known, the law of conditional probability may be use.

      This Law states that " the probability that both of two dependent events will occur is the probability of the first multiply by the probability that if the first has occurred the second will also happen."

       If two dependent events are  A and B then the probability that both of two dependent events A and B will occur is given as below-

        P (B) x P (A/B)

Where P (B) = Probability of occurring of event B ,

           P (A/B) = Probability that A happens given that B has happened.

                          ( Here probability of A conditioned by B )

TO SOLVE ABOVE LAW FOLLOWING MUST BE LEARN :-

( A ). Factorial Notation :

 1. n! = The product of first "n" natural numbers.

    

        n! = n x (n-1) x (n-2) x ... x 1

       n! = n x (n-1)!

       0! = 1

EXAMPLE :

1. 4! = 4 X 3 X 2 X 1 = 24

2. 5! = 5 X 4 X 3 X 2 X 1 = 120

3. 6! = 6 X 5 X 4 X 3 X 2 X 1 = 720 etc.

4. 6! = 6 x 5 x 4! or 6 x 5! etc.

2. If in a problem we select 'r' objects out of 'n' distinguishable objects        and arranging them in order, then we can notation as below :-

    

      ⁿP_r = n (n-1) (n-2) ... (n-r+1)  = n!/ (n-r)!

( B ). Combinations :

    " A combination is the number of ways in which a certain number of articles are selected from a lot, without regard to order in which they are drawn."

If number of combinations 'n' objects taken 'r' at a time then,

ⁿC_r  = n! / ⟨ r! x (n-r)! ⟩

EXAMPLE :

1. Total number of ways to select 4 articles from a lot of 10.

         = ¹⁰C₄ =10!/ ⟨ 4! x (10-4)! ⟩

                 =10!/ ⟨ 4! x 6! ⟩

                 = (10 x 9 x 8 x 7 x 6! )/( 4 x 3 x 2 x 1 x 6!)

                 = (10 x 9 x 8 x 7)/(4 x 3 x 2 x1)

                 = 210

2. ⁶⁰C₅ = 60!/ ⟨ 5! x (60-5)! ⟩

            = 60!/ ⟨ 5! x 55! ⟩

 =(60 x59 x58 x57x56 x55! )/(5 x 4 x 3 x 2 x 1 x55! )

 = ( 60 x59 x58 x57x56 )/ ( 5 x 4 x 3 x 2 x 1 )

 = 5461512

3. A lot of 30 bulbs out of which 5 are defective, 4 times are drawn from it. Find the probability that none is defective.

Solution :

For first draw : 

The number of non defective bulbs = 30-5 = 25

Hence, the probability that the first item is non defective = 25/30.

For second draw :

Total remains bulbs after first draw = 30-1 = 29 bulbs

Total remains non-defective bulb after one draw = 25-1 = 24

Hence, the probability that second item is non-defective = 24/29

Similarly,

For third draw = 23/28,

For forth draw = 22/27

Therefore, the probability that 4 times drawn are non-defective -

    = 25/30 x 24/29 x 23/28 x 22/27 = 0.4616 = 46.16%

4. A lot of 30 articles contains 5 defectives. A sample of 6 selected at random from the lot. What are the respective probabilities to get 0,1,2,3,4,5 defective articles in the sample of 6.

Solution :

Here Lot Quantity = 30

No. of defective articles = 5

Hence, No. of non-defective articles = 30-5 = 25

Here, 

No of ways to select 6 articles from the lot of 30 = ³⁰C₆  

                                                                           _{= 30!/( 6! x 24! )}

_{(i). Probability of Zero defective P (0) :}

      _{Here we need to take sample from non-defective only.}

_Hence,

    _{P (0) =} ²⁵C₆ / ³⁰C₆ 

         = ⟨ 25! / (6! 21! ) ⟩ / ⟨ 30! / (6! 24! ) ⟩

            = ( 25! x 6! x 24! ) / ( 6! x 21! x 30! )

 =(25 x24 x23 x22 x21! x 24! )/(21! x 30 x29 x 28 x 27x 26 x 25 x 24! )

 = (25 x 24 x 23 x 22) / (30 x 29 x28 x27 x 26 x 25)

 = 0.0007101455

(ii). Probability to get one defective article only P (1) :

    " This condition possible if we get 1 defective article from 5 defective and 5 non-defective from 25 non-defective articles. "

Hence,

    Probability to get 1 defective article  P(1) = ( ⁵C₁ x ²⁵C₅ ) / ³⁰C₆ 

= ⟨ ( 5!/(1! x 4! ) x 25!/(5! x 20!) ⟩ / ⟨ 30!/ (6! x 24!) ⟩

= 0.4474

Similarly,

(iii). Probability to get 2 defective article only P (2) :

        = P (2) = ( ⁵C₂ x ²⁵C₄ ) / ³⁰C₆

          _{= 0.21304}

(iv). Probability to get 3 defective article only P (3) :

        = P (3) = ( ⁵C₃ x ²⁵C₃ ) / ³⁰C₆

(v). Probability to get 4 defective article only P (4) :

        = P (4) = ( ⁵C₄ x ²⁵C₂ ) / ³⁰C₆

(vi). Probability to get 5 defective article only P (5) :

        = P (5) = ( ⁵C₅ x ²⁵C₁ ) / ³⁰C₆

_{NOTE : Here P (0)+ P(1) + P(2) + P(3) + P(4) + P(5) = 1}

      

I look forward to your valuable suggestions and ideas. Please give me support to improve my efforts to give valuable knowledge among peoples.

PLEASE READ THIS LINK TO FIND MUCH MOTIVATION ABOUT REAL LIFE SUCCESS : 

WE CAN WIN (greatprafool.blogspot.com)

FOR GROW MENTAL POWER AND INCREASE MEMOTY / MARKS ; PLEASE WATCH THIS LINK AND FOLLOW RULES CONTINIOUSLY :--

Post: Edit (blogger.com)

Also watch this link to learn in easy steps about NUMBER AND NUMBER SYSTEM , KNOW ABOUT DEVISIBILITY RULES :-

Post: Edit (blogger.com)

         

         

    

    

        

    

    

    

        

    

                       

MATHEMATICS IN EASY STEPS : NUMBER AND NUMBER SYSTEM . KNOW ABOUT DEVISIBILITY RULES

        

Mathematics is a easy subject but maximum find them so hard to learn and remember Mathematics functions and rules. That all because we not give time to understand basics of Mathematics rule. So learn here the basic rules step by step.

The first step to find understanding is learn about number and number system. Also divisibility rules so useful in solve questions. We find us uncomfortable if we not know as which number original number can be divide , that can be make easy if we know divisibility rules. Here all will describe in short notes so you can understand soon.

1. Types of Numbers :

Following are the basic types of numbers :-

1. Natural Numbers  (N) = (1,2,3,4,...)

2. Whole Numbers  (W) = (0,1,2,3,4,...)

3. Integers  (Z) = ( … , -3,-2,-1,0,1,2,3 , … )

4. Real Numbers  (R) = (23.057 , 0.057 , 0.5555 , 5/7 ,  √3 , -15 , 20 etc.)

5. Rational Numbers  (Q) = ( 1/2 , -3/5 , 2/9 , 0/2 , 99/100 , 1.56 etc.)

6. Irrational Numbers  (P) = (√2 , √3 , √5 , 0.343434 , Π , etc.)

7. Complex Numbers ⟨ ( a+b i )form ⟩ =( 4+5i , -3+4i , 2+√2i etc. )

other related numbers are :-

8. Imaginary Numbers

9. Discrete and Continuous Numbers

10. Prime Numbers

11. Composite Numbers

1. Natural Numbers  (N) :

    These are positive integers. Zero (0) not a natural number.

     (N) = (1,2,3,4,...)

2. Whole Numbers  (W) :

    These are the set of natural numbers with add "Zero" (0) .

    (W) = (0,1,2,3,4,...)

3. Integers  (Z) :

    These are the set of whole numbers with included negative natural numbers.

    (Z) = ( … , -3,-2,-1,0,1,2,3 , … )

4. Real Numbers  (R) :

    These are the numbers which can be use decimal also. Fractions also write in decimal form. It also includes all the irrational numbers such as  Π , √2 , √3 etc. 

All integers are real numbers but not all real numbers are integers.

Each real Numbers can be represented on number line.

Real numbers are combination of all numbers such as rational , irrational , decimal etc.

Real numbers includes all the positive and negative integers , whole numbers , fractions , repeating decimals , terminating decimals etc. complex numbers and imaginary numbers are not real numbers.

     (R) = (23.057 , 0.057 , 0.5555 , 5/7 ,  √3 , -15 , 20 etc.)

5. Rational Numbers  (Q) :

    These are the fractions where numerator or denominator numbers are integers that includes both positive and negative fractions but denominator never be zero (0) , but numerator can be zero.

Decimal numbers can be represented in fractional form.

Example : 0.256 , -256/1000 , -35/100 etc.

    (Q) = ( 1/2 , -3/5 , 2/9 , 0/2 , 99/100 , 1.56 etc.)

6. Irrational Numbers  (P) :

    These are the numbers which can't be possible to express in the form of P/Q (Fractional form).

    (P) = (√2 , √3 , √5 , 0.343434 , Π , etc.)

7. Complex Numbers ⟨ ( a+b i )form ⟩ :

    The numbers which in the form of ( a + b i ) is called complex numbers ; where "a" and "b" are real number and " i '' is an imaginary number.

Example : 4+5i , -3+4i , 2+√2i etc. are complex numbers.

8. Imaginary Numbers :

    Imaginary numbers are use in complex numbers . The square root of negetive number is represented by letter " i ".

Example : √(-2) , √-5 etc.

# SOLUTION OF IMAGINARY NUMBER :

    If X = √ (-81) then we can solve that as below -

        X = √ ( 81 x -1) = √81 x √(-1) = 9 x √(-1)

            = 9i

Hence X = 9i

9. Discrete and Continuous Numbers :

    The Natural numbers , Whole numbers , Integers , Rational numbers are referred to as discrete numbers. Discrete numbers can be countable.

But,

Real numbers can't be counted and are continuous numbers.

10. Prime Numbers :

    These are the natural numbers which have no factors other then 1 and itself. 1 is not a prime number. The smallest prime number is 2.

    Example : 2,3,5,7,... etc. are prime numbers.

11. Composite Numbers :

     Composite Numbers have some factors of a respective numbers such as 4, 15 are composite numbers because 4 is divisible by 2 ; similarly 15 is divisible by 3 and 5.

2. NUMBER SYSTEM :

A number system is defined as a system of writing to express numbers.

Number system is the mathematical representation of numbers of a given set of digits or other symbols in a systematic manner.

Following are the basic number systems :--

1. Decimal Number System

2. Binary Number System

3. Octal Number System

4. Hexadecimal Number System

In the number system , each number is represented by its base.

If the base is 2 , its called binary number system ;

If the base is 8 , its called octal number system ;

If the base is 10 , its called decimal number system ;

If the base is 16 , its called hexadecimal number system.

DECIMAL NUMBER SYSTEM :--

    In decimal number system we use the base 10 . It also called base 10 number system. Use numbers 0,1,2,3,4,5,6,7,8,9.

For representing decimal system we follow this rule ; for example :-

(342)₁₀ = ( 3x10² )+(4x10¹)+(2x10⁰)

             = 300+40+2

(348.258)₁₀ = ( 3x10² )+(4x10¹)+(8x10⁰)+( 2x10^-1 )+( 5x10^-2 )+( 8x10^-3 )

                     = 300+40+8+0.2+0.05+0.008

3. DIVISIBILITY RULES (DIVISION RULES) :

(1). Divisibility by 2 :

        If the ones digit of a number is either 0 or even number , then the given number will be divisible by 2.

Example : 2, 14, 246 , 298 etc.

(2). Divisibility by 3 :

        If the sum of all digits of number is divisible by 3 , then the given number will also be divisible by 3.

Example : 1608 ; Here 1+6+0+8 = 15 ; which is divisible by 3 so 1608 also divisible by 3.

(3). Divisibility by 4 :

       If the given number's tens and ones digit joining number is divisible by 4 , then given number will be divisible by 4.

Example : 954948 ; Here last 2 digits (48) is divisible by 4 so 954948 is divisible by 4.

(4). Divisibility by 5 :

       If the ones digit of a number is either 0 or 5 , then the given number will be divisible by 5.

Example : 485 , 590 are divisible by 5.

(5). Divisibility by 7 :

       Substract 2 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 7 , then the given number will be divisible by 7.

Example : 321468 → 32146 - (2x8) = 32130→3213-(2x0)=3213→

                                  321-(2x3)=315→31-(2x5)=21; 

                  Which is divisible by 7 ; so 321468 also be divisible by 7

(6). Divisibility by 8 :

       If the given number's hundreds, tens and ones digit joining number is divisible by 8 , then given number will be divisible by 8.

Example : 433872 ; Here last 3 digits (872) is divisible by 8 so 433872 is divisible by 8.

(7). Divisibility by 9 :

        If the sum of all digits of number is divisible by 9 , then the given number will also be divisible by 9.

Example : 4085289 ; Here 4+0+8+5+2+8+9 = 36 ; which is divisible by 9 so 4085289 also divisible by 9.

(8). Divisibility by 11 :

       If the difference between the sum of even numbers and odd numbers are either 0 or divisible by 11; then the give number will be divisible by 11.

Example : 50525827 ; Here start from ones the sum of even numbers = 2+5+5+5 = 17 and sum of odd numbers= 7+8+2+0 = 17 

Here, 

Sum of even no. - Sum of odd no. = 17-17 = 0 

So, given number will be divisible by 11.

(9). Divisibility by 13 :

       Add 4 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is divisible by 13 , then the given number will be divisible by 13.

Example : 5161 → 516 + (4x1) = 520→52+(4x0)=52→The result is divisible by 13 , so the original number will be divisible by 13.

(10). Divisibility by 17 :

       Substract 5 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 17,then the given number will be divisible by 17.

Example : 94843 → 9484 - (5x3) = 9469→946-(5x9)=901→90-(5x1)                                 =85 ; The result is divisible by 17 , so the original number will be divisible by 17.

(11). Divisibility by 19 :

       Add 2 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is divisible by 19,then the given number will be divisible by 19.

Example : 1132871637 → 113287163+(2x7) =          113287177→11328717+(2x7)=11328731→1132873+(2x1)                    =1132875→113287+ (2x5)=113297→11329+(2x7)= 11343→1134+(2x3)=1140→114+(2x0)=114→11+(2x4)=19  ; The result is divisible by 19 , so the original number will be divisible by 19.

(12). Divisibility by 23 :

       Add 7 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is divisible by 23 ,then the given number will be divisible by 23.

Example : 1207339 → 120733+(7x9) =120796→12079+(7x6)=12121→1212+(7x1)=1219→121+ (7x9)=184→18+(7x4)= 46 ; The result is divisible by 23 , so the original number will be divisible by 23.

(13). Divisibility by 29 :

       Add 3 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is divisible by 29,then the given number will be divisible by 29.

Example : 45327 → 4532+(3x7) =4553→455+(3x3)=464→

46+(3x4)=58 ; The result is divisible by 29 , so the original number will be divisible by 29.

(14). Divisibility by 31 :

       Substract 3 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 31,then the given number will be divisible by 31.

Example : 473153 → 47315-(3x3) =47306→4730-(3x6)=4712→

471-(3x2)=465→46-(3x5)=31 ; The result is divisible by 31 , so the original number will be divisible by 31.

(15). Divisibility by 37 :

       Substract 11 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 37,then the given number will be divisible by 37.

Example : 97791 → 9779-(11x1) =9768→976-(11x8)=888→

88-(11x8)=0 ; The result is 0 , so the original number will be divisible by 37.

(16). Divisibility by 41 :

       Substract 4 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 41,then the given number will be divisible by 41.

Example : 21730 → 2173-(4x0) =2173→217-(4x3)=205→

20-(4x5)=0 ; The result is 0 , so the original number will be divisible by 41.

(17). Divisibility by 43 :

       Add 13 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is divisible by 43,then the given number will be divisible by 43.

Example : 10750 → 1075+(13x0) =1075→107+(13x5)=172→

17+(13x2)=43 ; The result is divisible by 43 , so the original number will be divisible by 43.

(18). Divisibility by 47 :

       Substract 14 times the last digit from remaining number . Repeat the step as necessary up to 2 digit number. If result is either 0 or divisible by 47,then the given number will be divisible by 47.

Example : 9635 → 963-(14x5) =893→89-(14x3)=47; The result is divisible by 47 , so the original number will be divisible by 47.

I look forward to your valuable suggestions and ideas. Please give me support to improve my efforts to give valuable knowledge among peoples.

PLEASE READ THIS LINK TO FIND MUCH MOTIVATION ABOUT REAL LIFE SUCCESS : WE CAN WIN (greatprafool.blogspot.com)

FOR GROW MENTAL POWER AND INCREASE MEMOTY / MARKS ; PLEASE WATCH THIS LINK AND FOLLOW RULES CONTINIOUSLY :--

Post: Edit (blogger.com)

Also watch this link to learn in easy steps about PROBABILITY AND ITS APPLICATIONS  ( WITH SIMPLE EXAMPLES ) :-

Post: Edit (blogger.com)