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Principles of Statistics

Principles of Statistics. Chapter 2 Elements of Probability. Basic Concepts: Statistical Experiment : Process of collecting data Sample Space (S) : Collection of all outcomes of the experiment. Event : Any subset of the sample space Probability of the Event : Measure on the event.

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Principles of Statistics

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  1. Principles of Statistics Chapter 2 Elements of Probability

  2. Basic Concepts: Statistical Experiment: Process of collecting data Sample Space (S): Collection of all outcomes of the experiment. Event: Any subset of the sample space Probability of the Event: Measure on the event

  3. Probability Concept: • There are different approaches to define probability. These approaches are: • Subjective probability: • The main idea of subjective probability is to let the probability of an event reflect the personal belief in the chance of the occurrence of the event. For example, one may use personal probability when evaluating the chance of rain tomorrow.

  4. (2) Statistical probability: probability is the relative frequency of occurrence of an event in a large number of trials. P(A)= limit n(A)/N as N gets large, where n(A) is number of elements in A and N the number of elements in S. (3) Classical probability: If the sample space of an experiment consists of finite sample points which are equally likely, then for any event we define the probability of as: P(A)=n(A)/N, where n(A) is the number of outcomes in A and N is the number of outcomes in S.

  5. Example: Experiment of tossing a coin twice. S={HH, HT, TH, TT} A=getting at least one head P(A)=3/4=0.75 Example: Experiment ofthrowing two dice S={(1,1),(1,2), . . ., (1,6) (2,1), (2,2), . . ., (2,6) . . . (6,1), (6,2), . . ., (6,6)} A=sum of the points=7 - - -> P(A) = 6/36=1/6

  6. Probability Techniques: • Many problems in probability theory can be solved by simply counting the number of different ways that a certain event occurs. Combinatorial methods play an important role in computing probabilities whenever equally likely outcomes can be identified and sample space is finite. Now, we give a review of some counting methods. • A very basic technique that is frequently used in combinatorial problems is called the Multiplication principle which states that if a first operation can be performed in any ofn1 ways, and and a second operation can be performed in any ofn2 ways then both operations can be performed in (n1 n2) ways.

  7. Example:A die is rolled 5 times. Compute the probability that no two dice show the same number of spots. Solution:The sample space for this experiment is: S={(x1,x2,x3,x4,x5): xi=1,2,3,4,5,6, i=1,2,3,4,5} n(S)=6*6*6*6*6=(6^5) n(A)=6*5*4*3*2 Probability of A =Probability of no two dice have the same number = n(A)/n(S) = 0.093

  8. Assume that we have an urn containing balls of different colors. We are interested in drawing a sample from this urn. There are different procedures to obtain a sample. We will describe some of these procedures. (1) The balls may be drawn either sequentially, that is, one at a time, or simultaneously, that is, all at a time. (2) If the balls are drawn sequentially, then they may be drawn with replacement or without replacement. In this case we obtain an ordered sample, that is, there is a first ball, a second ball, and so on. (3) If the sampling is done simultaneously, then we obtain an unordered sample.

  9. A permutation is an ordered arrangement of all or part of a set of objects. The number of permutations of n distinguishable or different objects is n!=n(n-1)(n-2) . . . (3)(2)(1) Note that 0! = 1. • One may also be interested in the number of permutations of distinguishable objects taken r at a time. That is an ordered arrangements of r of a set n distinguishable objects. The number is called n permutation r (denoted by nPr) nPr= n(n-1) . . . (n-r+1) = n!/(n-r)!

  10. Example:There are 10 students in a class. What is the probability that no two students in the class have the same birthday? Assuming that each student in the class can have as his birthday any one of the 365 days in the year, so we ignore the existence of leap year, and that each day of the year is equally likely to be the person's birthday. Let A be the event that no two students in the class have the same birthday. Then P(A)= n(A)/n(S) = (365)(364) . . . (356)/(365^10)

  11. Example:Two balls are drawn without replacement from an urn containing 6 balls, of which 4 are white and 2 are red. Find the probability that: • both balls will be white • (b) both balls will be of the same color • (c) both balls will be of different color • (d) at least one of the balls will be white

  12. Solution: • Let A: two balls are white. • P(A)=4P2/6P2=(4)(3)/6(5) • = 12/30 • = 0.4 • (2)Let B be the event that both balls drawn are red. Then • P(B) =2P2/6P2 = 2/30 • = 0.067 • Hence, • P(both balls will be of the same color) = P(A) + P(B) • = 0.4 + 0.067 = 0.467

  13. (3) Let C be the event that both balls drawn are of different color. Then P(C) = [(2)(4)+(4)(2)]/30 = 16/30 = 0.533 (4) Let D be the event that at least one of the balls drawn will be white. Note that D = complement of B P(D)= 1 – P(B) =1 – 2/30 = 28/30 = 0.933

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