Geometric probablity
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The document discusses the history and development of probability theory. It begins by describing how the foundations of probability were established in the 17th century by Blaise Pascal and Pierre de Fermat through their study of games of chance. It then provides definitions and explanations of key probability concepts like sample space, events, and the probability of an event occurring. The document goes on to discuss different models for representing probability problems geometrically, including linear models, area models, and coordinate geometry models. It also covers famous probability puzzles and paradoxes like the Monty Hall problem, Bertrand's paradox, and the Sleeping Beauty problem. Finally, it briefly mentions some applications of probability in fields like quantum mechanics, polymer sciences, and
Geometric probablity
- 1. By- Harsha Hegde Karthik N R Manoj Hegde
- 2. A Time Travel into Probability Two French mathematicians: Blaise Pascal (1623–1662) Pierre de Fermat (1601–1665) found the mathematics of probability in the middle of the seventeenth century. Their first discoveries involved the probability of games of chance with dice and playing card s .
- 5. Introdoction to probability Probability is ordinarily used to describe an attitude of mind towards some proposition of whose truth we are not certain.The certainty we adopt can be described in terms of a numerical measure and this number, between 0 and 1, we call probability. The higher the probability of an event, the more certain we are that the event will occur. used widely in areas of study as mathematics, statistics, finance, gambling, science, artificial intelligence/machine learning and philosophy to, for example, draw inferences about the expected frequency of events.
- 6. Probability of occurrence of an event P(n)=No of favourable outcomes Sample Space(all possible outcomes)
- 7. Geometric Probability Modern probability theory is full of formulas and applications to modern events that are far removed from games of chance. Instead of using algebraic formulas to solve these probability problems, we use geometric figures. Geometric probability involves the distributions of length, area, and volume for geometric objects under stated conditions. These geometric figures allow you to picture the probabilities of a situation before solving the problem. This will help you develop a sense of a reasonable solution before you solve the problem.
- 9. Simple Problem Simulation Imagine that you are throwing a dart at a 10x10 square with a 4x4 square centrally placed. Then what is the probability of you hitting the dart into the inner block.?!
- 10. Favourable region=16 sq units Feasible region=100 sq units P= Area of favourable region =0.16 Feasible region Thus you fancy a 16% chance of hitting the target! This simple problem shows how logical is geometrical probability compared to classical approach.
- 12. As is evident from the previous slide if the area of the favorable region is increased consequently the probability of the favorable event increases .Simple but a very important conclusion!!
- 13. Various models in GP Linear Models Area Models Co-ordinate geometry models 3D models
- 14. Linear Models Example 1: A is driving one car in a line of cars, with about 150 feet between successive cars. Each car is 13 feet long. At the next overpass, there is a large icicle. The icicle is about to crash down onto the highway. If the icicle lands on or within 30 feet of the front of a car, it will cause an accident. What is the chance that the icicle will cause an accident?
- 16. The chance that the icicle will strike a passing automobile may be represented by line segments. The sample space is the distance from the front of one car to the front of the next car. The feasible region is A 163 feet B the 30 feet given in the problem plus the length of the car (13 feet). Sample space Both of these may be represented as segments Segment AB represents the sample space. Sample space = 150 feet (distance between cars) + 13 feet (length of a car) = 163 feet The feasible region may be represented by segment CD. Measure of feasible region = length of segment CD = 43 feet
- 17. Probability=43/163=0.264 The chance that the icicle will cause an accident is less than 30%!
- 18. Area Models Example 2:The ARIZONA CRATER A current scientific theory about the fate of dinosaurs on the earth suggests that they were made extinct by the effects of a large meteor which struck the earth in the Caribbean Sea, off the coast of Central America. The United States has been struck by large meteors in the past, as shown by the crater in Arizona at left. What is the chance that a meteor striking the earth will land in the United States?
- 20. The probability relationships for this problem may be represented by areas. The sample space is the surface area of the earth, or 196,940,400 square miles. The feasible region is the area of the United States, or 3,679,245 square miles.
- 21. Co-ordinate Geometry Models Another model we can use to represent probability problems is based on geometric figures placed on a coordinate grid. The problem in this section demonstrates how a coordinate system may be used to solve probability problems.
- 22. Example A line segment is 8 inches long. Two points are put on the segment at random locations. What is the probability that the three segments formed by the two points will make a triangle? This problem really is tedious and cumbersome in the classical method..See for yourself how simplified it is now!
- 24. What is the probability that segment AB, segment BC, and segment CD will make a triangle? This problem may be solved by using a coordinate system. let the length of segment AB = x. let the length of segment BC= y. The length of segment CD = 8 - x - y. The length of AB must be less than 8 inches, and the length of BC must be less than 8 inches. This may be represented on a coordinate system as shown in Figure.
- 26. In addition, the sum of the two segments also must be less than the total length of the original segment, so AB + BC is less than 8 inches. This may be written x + y < 8. This inequality is shown on the coordinate system in Figure .
- 28. The area of triangle QRS is the sample space. All the points in triangle QRS represent possible lengths for segment AB and segment CD. Point K, for example, has coordinates (6, 1). Point K represents a length of 6 for segment AB, and a length of 1 for segment Be The relationship between the three sides of a triangle is as follows: The sum of the lengths of any two sides of a triangle must be longer than the length of the third side. This results in the following inequalities, all of which must be true:
- 29. x + y > 8 - x - Y (AB + BC> CD) x + (8 - x - y) > Y (AB + CD > BC) y + (8 - x - y) > x (BC + CD > AB)
- 30. The first inequality may be simplified as follows: x+y>8-x-y 2x + 2y > 8 x + y> 4 (the shaded area in Figure ).
- 31. The second inequality may be simplified as follows: x + (8 - x - y) > y 8> 2y 4> y, or y < 4 (the shaded area shown in Figure ).
- 32. The third inequality may be simplified as follows: y + (8 - x - y) > x 8> 2x 4 > x, or x < 4 (the shaded area shown in Figure ).
- 33. The feasible region is the area of the common region defined by all three inequalities, as shown in Figure. The probability that the three segments will make a triangle is found by comparing the feasible region to the sample space: Feasible region = Area of Triangle EFG Sample space =Area of Triangle QRS
- 35. 3D models Probability in 3D models is a problem rejoiced in quantum physics. The normalization of the wave function done over a 3D space is the application of GP. ∫∫∫│w│2 dx dy dz w-wave function
- 36. Buffon’s Needle Experiment Georges Louis LeClerc, the Compte de Buffon (1707–1788), was a mathematician who is now famous for his needle experiment involving probability. Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips? He estimated the value of pi using this!(3.141592)
- 37. Bertrand’s paradox The Bertrand paradox goes as follows: Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle? Bertrand gave three arguments, all apparently valid, yet yielding different results.
- 38. The "random endpoints" method: The probability that a random chord is longer than a side of the inscribed triangle is 1/3.
- 39. The "random midpoint" method The probability a random chord is longer than a side of the inscribed triangle is 1/4.
- 40. The "random radius" method The probability a random chord is longer than a side of the inscribed triangle is 1/2.
- 41. Monty Hall problem The Monty Hall problem is a probability puzzle loosely based on the American television game show Let's Make a Deal and named after the show's original host, Monty Hall. It was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a) (Selvin 1975b). It became famous in the following form, as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):
- 42. Problem…. Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
- 43. In search of a new car, the player picks a door, say 1. The game host then opens one of the other doors, say 3, to reveal a goat and offers to let the player pick door 2 instead of door 1.
- 44. behind door 1 behind door 2 behind door 3 result if staying at door #1 result if switching to the door offered Car Goat Goat Car Goat Goat Car Goat Goat Car Goat Goat Car Goat Car The solution presented by vos Savant (1990b) The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.
- 45. Player picks car (probability 1/3) Host reveals either goat Switching loses. Player picks Goat A (probability 1/3) Host must reveal Goat B Switching wins Player picks Goat B (probability 1/3) Host must reveal Goat A Switching wins
- 46. The player has an equal chance of initially selecting the car, Goat A, or Goat B. Switching results in a win 2/3 of the time.
- 47. Two envelopes problem… You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.
- 48. Switching argument: I denote by A the amount in my selected envelope. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. The other envelope may contain either 2A or A/2. If A is the smaller amount, then the other envelope contains 2A. If A is the larger amount, then the other envelope contains A/2. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
- 49. So the expected value of the money in the other envelope is This is greater than A, so I gain on average by swapping. After the switch, I can denote that content by B and reason in exactly the same manner as above. I will conclude that the most rational thing to do is to swap back again. To be rational, I will thus end up swapping envelopes indefinitely. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction. The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.
- 50. Common resolutionobserve that A stands for different things at different places in the expected value calculation, step 7 above In the first term A is the smaller amount while in the second term A is the larger amount To mix different instances of a variable in the same formula like this is said to be illegitimate, so step 7 is incorrect, and this is the cause of By definition in one envelope is twice as much as in the other. Denoting the lower of the two amounts by X, we write the expected value calculation as the paradox. Here X stands for the same thing in every term of the equation
- 51. we learn that 1.5X is the average expected value in either of the envelopes, hence there is no reason to swap the envelopes.
- 52. Sleeping Beauty problem… Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be wakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be wakened and interviewed on Monday only. If the coin comes up tails, she will be wakened and interviewed on Monday and Tuesday. In either case, she will be wakened on Wednesday without interview and the experiment ends.
- 53. Any time Sleeping Beauty is wakened and interviewed, she is asked, "What is your belief now for the proposition that the coin landed heads?"
- 54. Solutions… Thirder position: The thirder position argues that the probability of heads is 1/3 Halfer position: The halfer position argues that the probability of heads is 1/2 Phenomenalist position The phenomenalist position argues that Sleeping Beauty's credence is meaningless until it is attached to consequences
- 55. Applications Quantum Mechanics Polymer sciences Ship wreck investigation Deep sea divers Meteorologists Excavators