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16 Pages

### L6 Probability

Course: ANY 191, Spring 2012
School: MapĂșa Institute of...
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Word Count: 815

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of PROBABILITY Concepts Probability MATH 30 Probability and Statistics OBJECTIVES At the end of this lesson, the students are expected to : 1. compute for the probability of an event 2. identify various ways of solving for the probability of an event. PROBABILITY DEFINITION A function which assigns to each event E in the sample space S some real number r [0,1 ] Event Probability Sample Event 2s [0, 1] E { }...

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of PROBABILITY Concepts Probability MATH 30 Probability and Statistics OBJECTIVES At the end of this lesson, the students are expected to : 1. compute for the probability of an event 2. identify various ways of solving for the probability of an event. PROBABILITY DEFINITION A function which assigns to each event E in the sample space S some real number r [0,1 ] Event Probability Sample Event 2s [0, 1] E { } Impossible Event S = Sure Event 0<r< 1 0 1 EXAMPLE Suppose the sample space S = {a1, a2, a3}. Indicate the cases where we have an acceptable assignment of probabilities to the simple events: a. P({a1}) = 1/3, P({a2}) = , P({a3}) = 1/3 b. P({a1}) = 1/3, P({a2}) = -1/3, P({a3}) = 1 c. P({a1}) = 1/5, P({a2}) = 3/5, P({a3}) = 1/5 d. P({a }) = 1/2, P({a }) = 0, P({a }) = 1/2 EXAMPLE A person decides to take a vacation and has three places in mind: Banaue, Baguio, and Tagaytay. She is twice as likely to go to Banaue as to Baguio, and three times as likely to go to Tagaytay as to the Banaue. Find the probability that she vacations in Baguio. PROBABILITY PROPERTIES OF PROBABILITY 1.0 < P(A) < 1 2.P(S) = 1 3.If A and B are mutually exclusive events then P(A B) = P(A) + P(B) 4. If A1, A2, is a sequence of mutually exclusive events, then P(A1 A2 An) = P(A1) + P(A2) + + P(An) PROBABILITY THEOREMS RELATING TO PROBABILITY Theorem 1 If is the empty set, then P() = 0 Theorem 2 If A is the complement of an event A, then P(A) = 1 P(A) Theorem 3 If A and B are any two events, then P(A B) = P(A) P(A B) Theorem 4 If A and B are any two events, then P(A B) = P(A) + P(B) P(A B) For any events A, B, and C P(A B C)=P(A)+P(B)+P(C)P(AB)P(AC) P(B C)+ Example A card is drawn at random from a standard deck of cards. Find the probability that the card is a black card; a face card; a black card or a face card. Example Of 120 students, 60 are studying Mathematics, 50 are studying Filipino, and 20 are studying Mathematics and Filipino. If a student is chosen at random, find the probability that the student (a) is studying Mathematics or Filipino, (b) is studying neither Mathematics nor Filipino. Example A box contains 4 white balls and 5 red balls. If three balls are drawn from one box, find the probability that a. one ball is white b. no ball is white c. at least one ball is white d. at most one ball is white Empirical (A posteriori) Approach The ratio of the frequency of the occurrence of the event f(E), to the number of trials in the experiment, n. f E (E) P( ) = n The empirical approach to determining probabilities relies on data from actual experiments to determine approximate probabilities instead of the assumption of equal likeliness Example On a production line, 35 electric can openers were found to be defective out of the 2000 randomly selected can openers that were tested. What is the probability that a can opener is defective? P(defective) = 35 2000 As a decimal 7/400 = .0175 As a percent.0175 = 1.75% Example Mary rolled a single die 20 times. She rolled 4-1's, 1-2's,5-3's, 6-4's, 0-5's, and 4-6's. Calculate the empirical probability that Mary rolled an "even" number. Solution: First determine the number of rolls which landed on an "even" number: 1 - 2, 6 - 4's, and 4 - 6's A total of 11 rolls landed on an "even" number. P(even) = 11 20 As a decimal: .55 As a percent: 55% Example Jim participated in an archery contest and out of his 20 shots, he hit the "Bulls-eye" 7 times. Based on these statistics, what is the chance that Jim will NOT hit the "Bulls-eye" on his next shot? Solution: If he hit the "bulls-eye" 7 out of 20 shots, he must have missed the "bulls-eye" 13 out of 20 shots. Therefore: P(NOT a bulls-eye) = 13 20 As a decimal 13/20 = .65 As a percent 13/20 = 65% Example We tossed a cup 25 times and recorded how it landed. It landed "up" 4 times, "down" 3 times and on it's "side" 18 times. If we tossed the cup 500 times how many times could we expect it to land "up", "down" or on it's "side"? Solution: The empirical probability for each possibility is: P(up) = 4/25 = .16 = 16% P(down) = 3/25 = .12 = 12% P(side) = 18/25 = .72 = 72% If we toss the cup 500 times simply multiply 500 by each decimal representation: So, P(up in 500 tosses) = (.16)(500) = 80 times! P(down in 500 tosses) = (.12)(500) = 60 times! P(side in 500 tosses) = (.72)(500) = 360 times! SUGGESTED READINGS TEXTBOOKS Beer, Ferdinand P., Johnston, E. Russell Jr., Mechanics for Engineers Statics, McGraw-Hill Book Company. Hibbeler, R. C., Engineering Mechanics Statics and Dynamics, 8th ed., Prentice Hall International, Inc., 1999 Meriam, J. L., Kraige, L. G., Engineering Mechanics Statics, John Wiley and Sons. Inc. Singer, Ferdinand L., Engineering Mechanics, Harper and Row, Ltd. WEBSITES: http://en.wikipedia.org/wiki/Mechanics http://www.personal.psu.edu/faculty/r/c/rce2/mcht111/111intro.htm
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