Probability Concepts Applications

Probability Concepts and Applications

Introduction Life is uncertain! We must deal with risk ! A probability is a numerical statement about the likelihood that an event will occur

Basic Statements About Probability The probability, P , of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1 . That is: 0  P(event)  1 2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1 .

Example Demand for white latex paint at Diversey Paint and Supply has always been 0, 1, 2, 3, or 4 gallons per day. (There are no other possible outcomes; when one outcome occurs, no other can.) Over the past 200 days, the frequencies of demand are represented in the following table:

Example - continued Quantity Demanded (Gallons) 0 1 2 3 4 Number of Days 40 80 50 20 10 Total 200 Frequencies of Demand

Example - continued Quant. Freq. Demand (days) 0 40 1 80 2 50 3 20 4 10 Total days = 200 Probability (40/200) = 0.20 (80/200) = 0.40 (50/200) = 0.25 (20/200) = 0.10 (10/200) = 0.05 Total Prob = 1.00 Probabilities of Demand

Types of Probability Objective probability: Determined by experiment or observation: Probability of heads on coin flip Probably of spades on drawing card from deck occurrences or outcomes of number Total occurs event times of Number ) (  event P

Types of Probability Subjective probability: Based upon judgement Determined by: judgement of expert opinion polls Delphi method etc.

Mutually Exclusive Events Events are said to be mutually exclusive if only one of the events can occur on any one trial

Collectively Exhaustive Events Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome: heads and tails as possible outcomes of coin flip

Example Outcome of Roll 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 Total = 1 Rolling a die has six possible outcomes

Example Outcome of Roll = 5 Die 1 Die 2 1 4 2 3 3 2 4 1 Probability 1/36 1/36 1/36 1/36 Rolling two dice results in a total of five spots showing. There are a total of 36 possible outcomes.

Probability : Mutually Exclusive P(event A or event B) = P(event A) + P(event B) or: P(A or B) = P(A) + P(B) i.e., P(spade or club) = P(spade) + P(club) = 13/52 + 13/52 = 26/52 = 1/2 = 50%

Probability: Not Mutually Exclusive P(event A or event B) = P(event A) + P(event B) - P(event A and event B both occurring) or P(A or B) = P(A)+P(B) - P(A and B)

P(A and B) (Venn Diagram) P(A) P(B) P(A and B)

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

Statistical Dependence Events are either statistically independent ( the occurrence of one event has no effect on the probability of occurrence of the other ) or statistically dependent ( the occurrence of one event gives information about the occurrence of the other )

Probabilities - Independent Events Marginal probability : the probability of an event occurring: [ P(A)] Joint probability : the probability of multiple, independent events, occurring at the same time P(AB) = P(A)*P(B) Conditional probability ( for independent events ) : the probability of event B given that event A has occurred P(B|A) = P(B) or, the probability of event A given that event B has occurred P(A|B) = P(A)

Probability(A|B) Independent Events P(B) P(A) P(A|B) P(B|A)

Statistically Independent Events 1. P(black ball drawn on first draw) P(B) = 0.30 ( marginal probability ) 2. P(two green balls drawn) P(GG) = P(G)*P(G) = 0.70*0.70 = 0.49 ( joint probability for two independent events ) A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball

Statistically Independent Events - continued 1. P(black ball drawn on second draw, first draw was green) P(B|G) = P(B) = 0.30 ( conditional probability ) 2. P(green ball drawn on second draw, first draw was green) P(G|G) = 0.70 ( conditional probability )

Probabilities - Dependent Events Marginal probability : probability of an event occurring P(A) Conditional probability ( for dependent events ) : the probability of event B given that event A has occurred P(B|A) = P(AB)/P(A) the probability of event A given that event B has occurred P(A|B) = P(AB)/P(B)

Probability(A|B) / P(A|B) = P(AB)/P(B) P(AB) P(B) P(A)

Probability(B|A) P(B|A) = P(AB)/P(A) / P(AB) P(B) P(A)

Statistically Dependent Events Assume that we have an urn containing 10 balls of the following descriptions: 4 are white (W) and lettered (L) 2 are white (W) and numbered N 3 are yellow (Y) and lettered (L) 1 is yellow (Y) and numbered (N) Then : P(WL) = 4/10 = 0.40 P(WN) = 2/10 = 0.20 P(W) = 6/10 = 0.60 P(YL) = 3/10 = 0.3 P(YN) = 1/10 = 0.1 P(Y) = 4/10 = 0.4

Statistically Dependent Events - Continued Then: P(L|Y) = P(YL)/P(Y) = 0.3/0.4 = 0.75 P(Y|L) = P(YL)/P(L) = 0.3/0.7 = 0.43 P(W|L) = P(WL)/P(L) = 0.4/0.7 = 0.57

Joint Probabilities, Dependent Events Your stockbroker informs you that if the stock market reaches the 10,500 point level by January, there is a 70% probability the Tubeless Electronics will go up in value. Your own feeling is that there is only a 40% chance of the market reaching 10,500 by January. What is the probability that both the stock market will reach 10,500 points, and the price of Tubeless will go up in value?

Joint Probabilities, Dependent Events - continued Then: P(MT) =P(T|M)P(M) = (0.70)(0.40) = 0.28 Let M represent the event of the stock market reaching the 10,500 point level, and T represent the event that Tubeless goes up.

Revising Probabilities: Bayes’ Theorem Bayes’ theorem can be used to calculate revised or posterior probabilities Prior Probabilities Bayes’ Process Posterior Probabilities New Information

General Form of Bayes’ Theorem

Posterior Probabilities A cup contains two dice identical in appearance. One, however, is fair (unbiased), the other is loaded (biased). The probability of rolling a 3 on the fair die is 1/6 or 0.166. The probability of tossing the same number on the loaded die is 0.60. We have no idea which die is which, but we select one by chance, and toss it. The result is a 3. What is the probability that the die rolled was fair?

Posterior Probabilities Continued We know that: P(fair) = 0.50 P(loaded) = 0.50 And: P(3|fair) = 0.166 P(3|loaded) = 0.60 Then: P(3 and fair) = P(3|fair)P(fair) = (0.166)(0.50) = 0.083 P(3 and loaded) = P(3|loaded)P(loaded) = (0.60)(0.50) = 0.300

Posterior Probabilities Continued A 3 can occur in combination with the state “fair die” or in combination with the state ”loaded die.” The sum of their probabilities gives the unconditional or marginal probability of a 3 on a toss: P(3) = 0.083 + 0.0300 = 0.383 . Then, the probability that the die rolled was the fair one is given by:

Further Probability Revisions To obtain further information as to whether the die just rolled is fair or loaded, let’s roll it again. Again we get a 3. Given that we have now rolled two 3s, what is the probability that the die rolled is fair?

Further Probability Revisions - continued P(fair) = 0.50, P(loaded) = 0.50 as before P(3,3|fair) = (0.166)(0.166) = 0.027 P(3,3|loaded) = (0.60)(0.60) = 0.36 P(3,3 and fair) = P(3,3|fair)P(fair) = (0.027)(0.05) = 0.013 P(3,3 and loaded) = P(3,3|loaded)P(loaded) = (0.36)(0.5) = 0.18 P(3,3) = 0.013 + 0.18 = 0.193

Further Probability Revisions - continued

To give the final comparison: P(fair|3) = 0.22 P(loaded|3) = 0.78 P(fair|3,3) = 0.067 P(loaded|3,3) = 0.933 Further Probability Revisions - continued

Probability Concepts Applications

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Probability Concepts Applications