SOUTHERN BICOL COLLEGESMABINI ST. MASBATE CITY

COMPILATION OF THE DIFFERENT

KINDS OF PROBABILITY

(PROBABILITY) GERLYN MAE C. ESPENILLA BSED-III (Major in Math)

Engr. MARIA ROMINA P. ANGUSTIA INSTRUCTOR

TABLE OF CONTENTS

*History of Probability 1 *Complementary Probability 2 *Joint Probability 3

-Mutually Exclusive Event -Non-Mutually Exclusive Event 4

*Conditional Probability 5 -Dependent Probability -Independent Probability 6

*Repeated Trial Probability 7

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Brief History of Probability

Concepts of probability have been around for thousands of years, but probability theory did not arise as a branch of mathematics until the mid-seventeenth century. During the fifteenth century several probability works emerged.  Calculations of probabilities became more noticeable during this time period even though mathematicians in Italy and France remained unfamiliar with these calculation methods (David, 1962).   In 1494, Fra Luca Paccioli wrote the first printed work on probability, Summa de arithmetica, geometria, proportioni e proportionalita (David, 1962).  In 1550, Geronimo Cardano inspired by the Summa wrote a book about games of chance called Liber de Ludo Aleae which means A Book on Games of Chance (David, 1962). 

     In the mid-seventeenth century, a simple question directed to Blaise Pascal by a nobleman sparked the birth of probability theory, as we know it today.  Chevalier de Mere gambled frequently to increase his wealth.  He bet on a roll of a die that at least one 6 would appear during a total of four rolls.  From past experience, he knew that he was more successful than not with this game of chance. Tired of his approach, he decided to change the game.  He bet that he would get a total of 12, or a double 6, on twenty-four rolls of two dice. Soon he realized that his old approach to the game resulted in more money.  He asked his friend Blaise Pascal why his new approach was not as profitable.  Pascal worked through the problem and found that the probability of winning using the new approach was only 49.1 percent compared to 51.8 percent using the old approach (Smith, 1996).

     This problem proposed by Chevalier de Mere is said be the start of famous correspondence between Pascal and Pierre de Fermat. They continued to exchange their thoughts on mathematical principles and problems through a series of letters.  Historians think that the first letters written were associated with the above problem and other problems dealing with probability theory.  Therefore, Pascal and Fermat are the mathematicians credited with the founding of probability theory (David, 1962).

     The topic of probability is seen in many facets of the modern world. The theory of probability is not just taught in mathematics courses, but can be seen in practical fields, such as insurance, industrial quality control, study of genetics, quantum mechanics, and the kinetic theory of gases (Simmons, 1992).

Complementary Events

- are two outcomes of an event that are the only two possible outcomes.

FORMULA :

P(A) + P(not A) = 1

P(not A) = 1 – P(A)

P(A) = 1 – P(not A)

Example

1: A die is rolled. Find the probability of not getting a 5.

Solution: We first find the probability of getting a 5.

P (getting a 5) = 16.

Hence the probability of not getting a five is the complementary event of getting a 5. So the probability is given by:

P (not getting 5) = 1 – P (getting a 5) = 1 – 16 = 56.

2. Out of the six balls given in a bag three are given to be white. Find the probability of not getting a white ball.

Solution: Probability of getting a white ball = P (white) = 36 = 12.

Hence the probability of not getting a white ball = 1 – 12 = 12.

Joint probability

- is a measure of two events happening at the same time, and can only be applied to situations where more than one observation can be occurred at the same time.

1.Mutually exclusive events - are two or more outcomes of an event that cannot occur at the same time.

FORMULA:

P(A or B) = P(A) + P(B) P(A and B) = 0

EXAMPLE :

As if is a fast bowler playing a inter-college cricket match. What is the probability of him getting 5 wickets or 2 wickets in the match when there are 20 percent chances of him getting 5 wickets and 60 percent chances of him getting 2 wickets?

Solution:

Asif can either get 5 wickets or he can get 2 wickets. Hence, the probability of him getting 5 wickets AND 2 wickets is zero.

Probability of him getting 5 wickets, P(A)=0.2

Probability of him getting 2 wickets, P(B)=0.6

Probability of him getting 5 wickets or 2 wickets, P(A or B)=P(A)+P(B)=0.2+0.6=0.8.

NON-MUTUALLY EXCLUSIVE

-Two events are called not mutually exclusive if they have al least one outcome common between them.

FORMULA :

P(A or B ) = P(A) + P(B) – P(A and B)

EXAMPLE:

A lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket is drawn at random, what is the probability that the number drawn is a multiple of 3 or 5?

SOLUTION: Let X be the event of ‘getting a multiple of 3’ and,Y be the event of ‘getting a multiple of 5’ The events of getting a multiple of 3 (X) = {3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48}Total number of multiple of 3 = 16P(X) = 16/50 = 8/25 The events of getting a multiple of 5 (Y) = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}Total number of multiple of 3 = 16P(X) = 10/50 = 1/5 Between the events X and Y the favorable outcomes are 15, 30 and 45.Total number of common multiple of both the number 3 and 5 = 3The probability of getting a ‘multiple of 3’ and a ‘multiple of 5’ from the numbered 1 to 50 = P(X ∩ Y) = 3/50Therefore, X and Y are non mutually exclusive events.We have to find out Probability of X union Y.So according to the addition theorem for mutually non- exclusive events, we get;

P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y)

Therefore, P(X U Y) = 8/25 + 1/5 - 3/50

= (16 + 10 -3)/50= 23/50

Hence, probability of getting multiple of 3 or 5 = 23/50

CONDITIONAL PROBABILITY

-the probability of an event ( A ), given that another ( B ) has already occurred .

2 types of Conditional Probability1. DEPENDENT PROBABILITY

-two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.

FORMULA :

P(A and B) = P(A) . P(B/A)

Example : At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68. What is the probability that a student takes Spanish given that the student is taking Technology?

Solution: P(Spanish|Technology)   = P(Technology and Spanish)  =

0.087   = 0.13  = 13%

P(Technology) 0.68

2. Independent Events

-An event which remains unaffected by previous event or set of events is known as an independent event.

P(A and B)=P(A)∗P(B)EXAMPLE :

On a library shelf, three geometry and five algebra books. Books are not replaced after someone borrow it. If two books are taken then what is the probability that first is of geometry and other of algebra.

Solution: The probability of first book to be of geometry = 38

The probability of second book to be of algebra = 57

As in total number of books one is already taken out, the total has become 7.

Probability of both events occurring = 38×57 = 1556

REPEATED TRIAL PROBABILITY

-a Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, "success" and "failure".

P=nC r ( p )r (q )n−r

EXAMPLE:

You are taking a 10 question multiple choice test. If each question has four choices and you guess on each question, what is the probability of getting exactly 7 questions correct?

SOLUTION:P=nC r ( p )r (q )n−r

n = 10r= 7n – r = 3p = 0.25 = probability of guessing the correct answer on a questionq = 0.75 = probability of guessing the wrong answer on a question

P=10 C 7¿

P= 405131072 = 3.089904785x10−3 or 0.3099%