GOVERNMENT COLLEGE OF ENGINEERING CH. SAMBHAJINAGAR A Presentation on BASIC CONCEPTS OF PROBABILITY Presented by Nikhil Vishnu Dhakne (MT24F08F002) Under the guidance of Prof. Rohini ma’am CONTENTS • What is Probability? • Term used in Probability • Three Types of Probability • General probability rules WHAT IS PROBABILITY? • Probability is the measure of how likely an event is. • The probability of an event is a number between 0 and 1 that expresses how likely it is to occur. • “The ratio of the number of favourable cases to the number of all the cases“ or P(E) = Number of outcomes favourable to E_____ Number of all possible outcomes of the experiment Term used in Probability • Event – an event is one occurence of the activity whose probability is being calculated. • Outcome - an Outcome is one possible result of an event. • Success - an outcome that we want to measure. • Failure - an outcome that we don’t want to measure. • Sample Space - The collection of all possible outcomes is known as Sample Space. • Equally Likely Outcomes - In this 50-50% chance are there of an outcome. THREE TYPES OF PROBABILITY • Theoretical Probability is based on a mathematical model of probability. • Empirical Probability ( or experimental probability) is based on how often an event has occurred in the past. • Subjective Probability is based on a person’s belief that an event will occur. THEORETICAL PROBABILITY οWe write P(E) to denote the probability of the event E. π(πΈ) π πΈ = π(π) ππ’ππππ ππ π€ππ¦π π‘π πππ‘ π€βππ‘ π¦ππ’ π€πππ‘ = π‘ππ‘ππ ππ’ππππ ππ πππ π ππππππ‘πππ THEORETICAL PROBABILITY • Example: I have 3 blue candies, 5 red candies and 2 yellow candy. Find the probability of the red candies. • Solution: 5 ππ.ππ πππ • P(red) = 10 ππ.ππ πππ πππππππ (πππ π ππππππ‘πππ ) 1 = 2 EMPIRICAL PROBABILITY • Empirical ( or experimental ) probability is of an event is an "estimate" that the event will happen based on how often the event occurs after collecting data or running an experiment (in a large number of trials). # ππ π π’ππππ π ππ Empirical Probability = # ππ π‘πππππ Empirical probability only makes sense if the event is repeatable. SUBJECTIVE PROBABILITY • Subjective Probability measures a person’s belief that an event will occur . • These probabilities vary from person to person. • They also change as one learns new information. • No mathematical calculations or proof behind this types of probability but it could be illustrated the following way: π π₯ = ππππππ ππ ππππππ π‘βππ‘ π₯ ππ π‘ππ’π SUBJECTIVE PROBABILITY • Example: • In a car parking area there are totally 32 cars available. In those, 3 are Innova cars, 8 are Swift cars, 15 are Accent cars and remaining are tata cars. Compute the probability for choosing i) Swift cars ii) tata cars? • Solution: • Number of Innova cars = 3 Number of Swift cars = 8 Number of Accent cars = 15 • Solution: • Let A be the event of selecting Swift cars. So, n(A) = 8 = `8/32` = `1/4` Let B be the event of selecting tata cars. Number of tata cars = 32 – (3+8+15) = 6 So, n(B) = 6 = `6/32` = `3/16` GENEREL PROBABILITY RULES RULE 1 THE PROBABILITY OF AN IMPOSSIBLE EVENT IS 0 WHILE THE PROBABILITY OF A CERTAIN EVENT IS 1. THEREFORE, THE RANGE OF ALL POSSIBILITIES IS π ≤ π· π¨ ≤ π. Example: Six-Sided Die It is impossible to roll an eight on a six-sided die. Thus, π·(π«π¨π₯π₯π’π§π π) = π. It is certain that you will roll a number between 1 and 6 if you roll a six-sided die. Thus, P(any number between 1 and 6) = 1 RULE 2 The sum of all of the probabilities for possible events is equal to 1. Example: Cards In a standard 52-card deck there are 26 black cards and 26 red cards. All cards are either black or red. 26 26 P(red)+P(black)= + = 1 52 52 RULE 3 The complement of any outcome is equal to one minus the outcome. In other words: π(π΄π ) = 1 − π(π΄) It is also true then that π(π΄) = 1 − π(π΄π ) Example: Rain According to the weather report, there is a 30% chance of rain today: π(π πππ) = .30 Raining and not raining are complements. Therefore π·(π΅ππ ππππ) = π·(πΉππππ ) = π − π·(πΉπππ) = π−. π =. ππ Therefore, there is a 70% chance that it will not rain today. RULE 4 ADDITION RULE • Determining the probability that either one or both events occur depends on whether the events are mutually exclusive or not. This is also known as a union and is represented by ∪. • π·(π¨ ∪ π©) is read as “probability of A or B.” Note that this also includes the possibility of both A and B occurring at the same time. If two events, A and B, are mutually exclusive, then the probabilities of the two events can be added. In other words, π·(π¨ ∪ π©) = π·(π¨) + π·(π©). This rule extends beyond two events if all are mutually exclusive. For example, if A, B, and C are all mutually exclusive events, then π·(π¨ ∪ π© ∪ πͺ) = π·(π¨) + π·(π©) + π·(πͺ) EXAMPLE: KING OR ACE In a standard 52-deck of cards, what is the probability of randomly selecting a king or ace? 1 1 π ππππ = = .077 π(πππ) = = .077 13 13 These events are mutually exclusive because a card cannot be both a king and an ace. 1 1 π·(ππππ ∪ πππ) = + =. πππ+. πππ =. πππ 13 13 If two events, A and B, are NOT mutually exclusive, then the probability of A or B is equal to the probability of A and the probability of B minus the overlap of the two. In other words, π·(π¨ ∪ π©) = π·(π¨) + π·(π©) − π·(π¨ ∩ π©) The probability of A and B must be subtracted when there is overlap otherwise this area is counted twice. π·(π¨ ∪ π©) = π·(π¨) + π·(π©) − π·(π¨ ∩ π©) EXAMPLE: ICE CREAM A large-scale survey finds that: 90% of college students enjoy eating chocolate ice cream, 70% of college students enjoy eating mint chocolate chip ice cream, 65% of college student enjoy eating both chocolate and mint chocolate chip ice cream. What proportion of college students enjoy eating chocolate or mint chocolate chip ice cream? These events are NOT mutually exclusive because it was possible for a participant to enjoy both. π πΆβππππππ‘π ∪ ππππ‘πΆπΆ = π(πΆβππππππ‘π) + π(ππππ‘πΆπΆ) − π(πΆβππππππ‘π ∩ ππππ‘πΆπΆ = .90 + .70 − .65 = .95 The probability that a randomly selected college student will enjoy chocolate or mint chocolate chip ice cream is .95 RULE 5 MULTIPLICATION RULE Determining the probability that both events occur depends on whether the events are independent or not. This is known as an intersection and is represented by ∩. π(π΄ ∩ π΅) is read as “probability of A and B.” If two events, A and B, are independent, then the probability of A and B is equal to the product of the two. In other words π· π¨∩π© =π· π¨ ×π· π© . This rule extends beyond two events. For example, if A, B, and C are all independent of one another, then π·(π¨ ∩ π© ∩ πͺ) = π·(π¨) × π·(π©) × π·(πͺ) EXAMPLE: FLIPPING A COIN When flipping a fair coin (i.e., 50% heads and 50% tails) three times, what is the probability that you will flip heads all three times? π(π») = 1/2 = .5 π(π»π»π») = π(π») × π(π») × π(π») = .5 × .5 × .5 = .125 The probability of flipping heads all three times is .125 What is the probability of flipping heads, tails, tails, in that order? π(π») = 1/2 = .5 πππ π(π) = 1/2 = .5 π(π»ππ) = π(π») × π(π) × π(π) = .5 × .5 × .5 = .125 EXAMPLE: GENDER AND SPORTS PARTICIPATION A study by Sabo & Veliz (2008) surveyed 2,132 American children about their participation in organized and team sports. In the sample, 50.7% were boys and 49.3% were girls. Of the boys, 73% currently participate in sports. Of the girls, 63% currently participate in sports. Gender and sports participation are NOT independent. What is the probability that you would select a girl who currently participates in sports? π(ππππ ∩ π ππππ‘π ) = π(ππππ) × π(π ππππ‘π β£ ππππ) = .493 × .63 = .311 What is the probability that a randomly selected individual from this sample would be a boy who does not currently participate in sports? π(πππ¦ ∩ πππ ππππ‘π ) = π(πππ¦) × π(πππ ππππ‘π β£ πππ¦) = .507 × (1 − .73) = .507 × .27 = .137 RULE 6 CONDITIONAL PROBABILITY π· π¨ π© ππ ππππ ππ “ππππππππππ‘π¦ ππ π΄ πππ£ππ π΅. ” π·(π¨ β£ π©) = π·(π¨ ∩ π©) π·(π©) π·(π© β£ π¨) = π·(π¨ ∩ π©) π·(π¨) Clubs In a standard 52-card deck. What is the probability that a randomly selected card is a club given that it is a black card? π πππ’π = 1/4 = .25 π(πππππ) = 2/4 = .50 π(πππ’π ∩ πππππ) = 13/52 = .25 π(πππ’π ∩ πππππ) .25 π(πππ’π β£ πππππ) = = = .50 π(πππππ) .50 Given that a randomly selected card is black, there is a 50% chance that it's a club.
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