Probability 101 Compiled by Henry Posters

Introduction Living with uncertainty is part of everyday life. As probability (defined by Webster’s as: the number of times something is likely to occur over the range of possible occurrences) affects many things we do, we would like to able to quantify as best as we can the amount of uncertainty. For example, before stepping into an airplane, people who are afraid of flying should know that the probability of a significant incident is reassuringly minute (flight delays and lost luggage not included…). Vacationers bound for San Diego would like to know the likelihood of enjoying at least 20 sunny days in August (very likely as we all know). Dreamers who play the lottery ought to know the odds of winning the jackpot (extremely small) before buying lots of tickets. One can estimate these likelihoods using the theory of probability, which provides mathematical methods for quantifying chances/likelihoods of random events in order to predict the behavior of defined systems. So, with sufficient knowledge of the physics of failures an automotive engineer should be able to quantify the likelihood of driving a car 100,000 miles without experiencing any engine failures.

Rules and Properties of Probability 1.0 Sample Spaces and Sets In statistics the set of all possible outcomes of an experiment in a given situation is called the Sample Space (S). An Event is any collection (subsets) of outcomes that are contained in the sample space (S). See diagrams 1 and 2. Sets, and the relationship between them, can be best explained with the help of Venn diagrams shown on page 7. A Venn diagram is graphical representation of the relationship between multiple events. To simplify probability equations it is convenient to use the customary notations of sets. Ac denotes the event that A does not occur. It is called the compliment of event A. The event that either A or B occurs is called a union of A and B. It is denoted by A ∪ B. See Venn diagram 4. Its compliment (A ∪ B)c is shown in diagram 5. Similarly, the event that C and D both occur is written as B ∩ D. It is referred to as the intersection of C and D. It is shown in Venn diagram 6. If events have no points in common (they cannot both occur) they are called mutually exclusive. See Venn diagram 3.

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2.0 Relationships among Sets 2.1 The compliment of an event A, denoted as Ac, is the event that consists of all the outcomes (points) of the sample space that are not contained in A. In other words, Ac is the event that A will not happen. See Venn diagram 1. 2.2 The union of two events A and B, denoted by A ∪ B and read “A or B”, is the event consisting of all the outcomes (points) that are either in A or in B or in both events. The union includes outcomes for which both A and B occur as well as outcomes for which exactly one occurs. See Venn diagram 4. 2.3 The intersection of two events A and B, denoted by A ∩ B and read “A and B” is the event consisting of all outcomes (points) that are in both A and B. See Venn diagram 6. 2.4 Similarly, the intersection of three events A, B, and C is denoted by A ∩ B ∩ C It is the event that consists of all outcomes (points) that are in A and B and C. See diagram 9. 2.5 From Venn diagram 5 we can see that (A ∪ B)c is the same event as (A)c ∩ (B)c 2.6 Event A ∩ Bc and event B ∩ Ac are depicted in respectively diagram 7 and 8. 2.7 From Venn diagrams 10, 11 and 12 we can derive that: A ∪ B = A ∪ (B ∩ Ac) = (A ∩ Bc) + (A ∩ B) + (B ∩ Ac) 2.8 When A and B have no outcomes in common, they are said to be mutually exclusive or disjoint events. The events cannot happen at the same time. See Venn diagram 3.

3.0 Properties of Probability Given an experiment and a sample space (S), the objective of probability is to assign to each event (A) a number P(A) that will give a measure of the chance that event (A) will occur. The probability of an event is always a positive real number or zero. 3.1 Probability that event A will happen is P(A) = the number of ways event A can happen, divided by the total number of possible outcomes. n( A ) From this basic premise it is reasonable to expect that 0 ≤ P( A ) ≤ 1 S The smallest probability we can have is zero, which means that event A will never happen. The largest probability is 1. In that case event A will always happen. P(A ) =

Some examples:

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The probability that an impossible event (such as rolling 15 with two dice) will happen is 0. The probability of an event that must happen (such as flipping either a head or a tail) is 1. 3.2 The probability of a sample space is always 1. Symbolically: P(S) = 1 for any sample space. 3.3 For any event A, P(A) + P(Ac) = 1 and consequently P(A) = 1-P(Ac). In other words: The probability that an event will occur and the probability that it will not occur always add up to 1. Events that cannot possible occur have the probability of zero. 3.4 If event A and event B are mutually exclusive (cannot both happen at the same time), then P(A and B) = P(A ∩ B) = 0. 3.5 For any two events A and B, P(A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) When A and B are mutually exclusive then P(A ∪ B) = P(A) + P(B)

4.0 Conditional Probability 4.1 Conditional probability is defined as P(A|B) which is the probability of event (A) given that event (B) has occurred. For any two events (A) and (B) with P(B) > 0, the conditional probability of (A) given that (B) has occurred is:

P( A | B ) =

P( A ∩ B ) P( B )

Example of Conditional Probability: Suppose that of all individuals who purchase a Ford SUV, 60 percent select a model with a sunroof, 40 percent pick the same model with leather seats, and 30 percent select one that has both, a sunroof and leather seats. Now consider randomly selecting a buyer and let A = SUV with sunroof and B = SUV with leather seats. Then P(A) = 0.6, P(B) = 0.4 and P(A and B) = P(A ∩ B) = 0.3 Given that a selected individual bought a Ford SUV with leather seats, what is the probability that this car has also a sunroof? P( A ∩ B ) 0.3 = = 0.75 That is, of all those who bought a Ford SUV with leather P( B ) 0.4 seats, 75 percent got one that has also a sunroof. P( A | B ) =

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5.0 General Rule of Multiplication P( A ∩ B ) by P(B) we get: P( B ) P(A ∩ B) = P(B) • P(A|B) Consideration of P(B|A) gives P(A ∩ B) = P(A) • P(B|A)

By multiplying both sides of P( A | B) =

The general rule of multiplication states: The probability that two events A and B will both occur is the product of the probability that one of them will occur and the probability that the other event will occur given that the first has occurred. When events A and B are independent then P(A ∩ B) = P(A) • P(B) Independent means that the occurrence or non-occurrence of one event has no bearing on the chance that the other event will happen.

Probability in Quality and Reliability Engineering Example 1 A RADAR system has three sub-systems. Extensive testing of the three sub-systems has revealed the following probabilities of failure-free performance: Sub-system 1: P(no failures over five year period) = 0.99 Sub-system 2: P(no failures over five year period) = 0.97 Sub-system 3: P(no failures over five year period) = 0.95 If any of the three sub-systems fails, the entire radar system fails. Failure of one sub-system has no bearing on the chance that any of the other sub-systems will fail. What is the probability that the radar system will perform, failure free, for a period of five years? The probability that the radar system will perform failure-free for a five year time period is: P(zero failures Radar system) = P(zero failures subsystem1 ∩ zero failures subsystem2 ∩ zero failures subsystem3) = P(0 failures subsystem1) • P(0 failures subsystem2) • P(0 failures subsystem3) = (0.99) • (0.97) • (0.95) = 0.912 or 91.2 percent. Example 2 A modern airplane requires only one out of the three (or four) engines to successfully operate during flight. The chance that a well-maintained engine fails is very small. Let the probability of engine failure during a 14 hour flight be 0.5 percent. P(engine failure during 14 hour flight) = 0.005 What is the probability that an airplane, powered by four engines, will safely fly from Los Angeles to Sydney Australia? Flight time is 14 hours. The plane will not make it to Sydney if all four engines would fail. The probability of that happening is quite small. Using the multiplication rule we get: P(all four engines fail) = P(engine1 fails ∩ engine2 fails ∩ engine3 fails ∩ engine4 fails) = P(engine1 fails)•P(engine2 fails)•P(engine3 fails)•P(engine4 fails) = (0.005)4 = [5(10)-3 ]4 = 5(10)-12 = 0.000000000005

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There are only five combinations of engines working/failing possible. 1. All four engines work, none fails 2. Three engines work, one fails 3. Two engines work, two fail 4. One engine works, three fail 5. All four engines fail The sum of the probabilities of events one to five is 1. Hence, the probability that at least one engine works is 1 – P(all four engines fail) = 1 – 5(10)-12 = 1 – 0.000000000005 = 0.999999999995 or 99.9999999995 percent. The probability that a modern day airplane will safely complete a 14-hour flight is extremely high.

Probability in Tennis In tennis the most important (and most complex) shot is the serve. The objective of the server is to make it as difficult as possible for the opponent to return it (or not be able to return the ball at all). Most recreational tennis players hit a strong (50 to100 mph) first serve and a rather weak (less than 50mph) second one. For an opponent a weak serve offers a great opportunity to hit a winner. Let’s assume a probability of 60 percent that a player’s first (cannon ball) serve makes it over the net into the service box (we will add probability data about the opponent’s chance of returning it successfully later) and a 90 percent chance that the second (weak) serve will be in. What would be the chance of getting at least one serve in? Answer: The probability of hitting a successful first (strong) serve is P(good first serve) = 0.6 The probability of not hitting a good first serve is 1-P(good first serve) =1-0.6 = 0.4 The probability of hitting a successful second (weak) serve is P(good second serve) = 0.9 The probability of missing the second serve is 1-P(good second serve) = 1-0.9 = 0.1 Given that the player serves both the first and the second serve there can be only three outcomes. a) Both serves are out. b) The first serve is out and the second one is in. c) Both serves are in (in reality the player would serve only once). P(two in)+P(one in)+P(zero in) = 1 The general rule of multiplication tells us that the probability that two events A and B will both occur is the product of the probability that one of them will occur and the probability that the other event will occur given that the first has occurred. The multiplication rule is written as: P(A ∩ B) = P(A)•P(B|A). When events A and B are independent then P(A ∩ B) = P(A)•P(B) Independent means that the occurrence or non-occurrence of one event has no connection to the likelihood that the other event will happen. In the tennis example, the chance that the player will

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miss the second serve does not depend on whether or not he/she misses the first serve (however, if the server gets the fist serve in there is no need to serve a second one). The chance that the first serve and the second serve will be out is: P(A ∩ B) = P(A) • P(B) P(zero serve in) = (0.4)•(0.1) = 0.04 The chance of getting at least one serve in is 1 minus the chance that none will be in. P(at least one serve in) = 1 - 0.04 = 0.96 or 96 percent. Some tennis players never compromise on their second serve. They will hit the second serve as fast as the first one. In that case the probability of getting at least on serve in will be reduced to 1-(0.4)•(0.4) = 1-0.16 = 0.84 or 84 percent.

Probability in Decision Making Decision-making under risk (probabilistic decision-making) requires a basic understanding of the rules of probability. To make intelligent decisions or predictions about the quality & reliability of a business, product or service we must have useful information about the business, product or service in question. Such information is usually obtained from sampling, taking quantitative measurements, and converting sample data into information. Quality professionals must understand that uncertainty/risk is introduced every time data is used that have been derived from samples (the data do not always represent “certainty”). Consequently, a lack of understanding of the rules of probability may lead to incorrect decision-making.

References: John Freund, “Introduction to Probability” Sheldon Ross, “A First Course in Probability” Jay Devore, Probability and Statistics for Engineering and the Sciences” http://www.asqsandiego.org/ Library: “Taking a Sample”

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