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barcode reader c# source code i i =1 in C#.NET
i i =1 PDF417 Reader In Visual C#.NET Using Barcode recognizer for .NET framework Control to read, scan PDF 417 image in Visual Studio .NET applications. www.OnBarcode.comRecognize PDF 417 In C# Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications. www.OnBarcode.comProbability and Random Variables 145
Barcode Reader In Visual C# Using Barcode recognizer for Visual Studio .NET Control to read, scan barcode image in Visual Studio .NET applications. www.OnBarcode.comBar Code Scanner In Visual C# Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET applications. www.OnBarcode.comexample: In the experiment of flipping two coins, let the event A = obtain at least one head The sample space contains four elements ({HH, HT, TH, TT}) s = 3 because there are three ways for our outcome to be considered a success ({HH, HT, TH}) and f = 1 Thus P( A ) = 3 3 = 3+1 4 Read PDF 417 In C#.NET Using Barcode recognizer for .NET Control to read, scan PDF417 2d barcode image in Visual Studio .NET applications. www.OnBarcode.comDecoding PDF417 2d Barcode In .NET Using Barcode scanner for ASP.NET Control to read, scan PDF417 2d barcode image in ASP.NET applications. www.OnBarcode.comexample: Consider rolling two fair dice and noting their sum A sample space for this event can be given in table form as follows: PDF417 2d Barcode Reader In .NET Using Barcode reader for VS .NET Control to read, scan PDF417 image in Visual Studio .NET applications. www.OnBarcode.comPDF417 2d Barcode Recognizer In Visual Basic .NET Using Barcode recognizer for .NET framework Control to read, scan PDF417 2d barcode image in .NET applications. www.OnBarcode.comFace 1 2 3 4 5 6
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+ P (12) = 1 2 + + 36 36
1 = 1 36 Probabilities of Combined Events
P(A or B): The probability that either event A or event B occurs (They can both occur, but only one needs to occur) Using set notation, P(A or B) can be written P( A B) A B is spoken as, A union B P(A and B): The probability that both event A and event B occur Using set notation, P(A and B) can be written P( A B) A B is spoken as, A intersection B example: Roll two dice and consider the sum (see table) Let A = one die shows a 3, B = the sum is greater than 4 Then P(A or B) is the probability that either one die shows a 3 or the sum is greater than 4 Of the 36 possible outcomes in the sample space, there are 32 possible outcomes that are successes [30 outcomes greater than 4 as well as (1,3) and (3,1)], so P( A or B) = 32 36 146 U Step 4 Review the Knowledge You Need to Score High
There are nine ways in which a sum has one die showing a 3 and has a sum greater than 4: [(3,2), (3,3), (3,4), (3,5), (3,6), (2,3), (4,3), (5,3), (6,3)], so P( A and B) = 9 36 Complement of an event A: events in the sample space that are not in event A The complement of an event A is symbolized by A , or A c Furthermore, P( A ) = 1 P (A) Mutually Exclusive Events
Mutually exclusive (disjoint) events: Two events are said to be mutually exclusive (some texts refer to mutually exclusive events as disjoint) if and only if they have no outcomes in common That is, A B = If A and B are mutually exclusive, then P(A and B) = P ( A B ) = 0 example: in the twodice rolling experiment, A = face shows a 1 and B = sum of the two dice is 8 are mutually exclusive because there is no way to get a sum of 8 if one die shows a 1 That is, events A and B cannot both occur Conditional Probability
Conditional Probability: The probability of A given B assumes we have knowledge of an event B having occurred before we compute the probability of event A This is symbolized by P(AB) Also, P ( A  B) = P ( A and B) P (B ) Although this formula will work, it s often easier to think of a condition as reducing, in some fashion, the original sample space The following example illustrates this shrinking sample space example: Once again consider the possible sums on the roll of two dice Let A = the sum is 7, B = one die shows a 5 We note, by counting outcomes in the table, that P(A) = 6/36 Now, consider a slightly different question: what is P(AB) (that is, what is the probability of the sum being 7 given that one die shows a 5) solution: Look again at the table:

