Unit-2.pdf

# Unit-2.pdf - PROBABILITY Sercan Gr Vienna University of...

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PROBABILITY Sercan Gür Vienna University of Economics and Business Department of Statistics and Mathematics WU Vienna,12 March 2015

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1 Computing Probabilities: Counting Methods The Multiplication Principle Permutations and Combinations 2 Conditional Probability 3 Independence 2/33
Computing Probabilities: Counting Methods Methods of counting are used in order to construct probability assignments on finite sample spaces. Examples: Suppose that a fair coin is thrown twice and the sequence of heads and tails is recorded. The sample space is Ω = { hh , ht , th , tt } . We assume that all outcomes in Ω are equally likely, i.e. has probability, .25. Let A denote the event that at least one head is thrown. Then A = { hh , ht , th } , and P ( A ) = . 75. We will extend the previous example to more general sample space Ω . If there are N elements in Ω , each of them has probability, 1 / N . If A can occur in any of n mutually exclusive ways, then P ( A ) = n / N , or P ( A ) = number of ways A can occur total number of outcomes 3/33

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1 Computing Probabilities: Counting Methods The Multiplication Principle Permutations and Combinations 2 Conditional Probability 3 Independence 4/33
Theorem 1.1 (Multiplication Principle) If one experiment has of m outcomes and another experiment has n outcomes, then there are m × n possible outcomes for the two experiments. Proof. Denote the outcomes of the first experiment by a 1 , . . . , a m and the outcomes of the second experiment by b 1 , . . . , b n . The outcomes for the two experiments are the ordered pairs ( a i , b j ) . These pairs can be exhibited as ( a 1 , b 1 ) ( a 1 , b 2 ) · · · ( a 1 , b n ) ( a 2 , b 1 ) ( a 2 , b 2 ) · · · ( a 2 , b n ) . . . . . . . . . . . . ( a m , b 1 ) ( a m , b 2 ) · · · ( a m , b n ) m × n possible outcomes Examples: Playing cards have 13 face values and 4 suits. There are thus 4 × 13 = 52 face-value/suit combinations. 5/33

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Tree Diagram: A coin and a die are thrown at random. Find the probability of: I getting a head and an even number I getting a head or tail and an odd number Figure: Tree Diagram How many different meals can be made from four kinds of meat, six vegetables and three starches if a meal consists of one selection from each group? 6/33
Theorem 1.2 (Extended Multiplication Principle) If there are p experiments and the first has n 1 possible outcomes, the second n 2 , . . . , the pth n p possible outcomes, then there are a total of n 1 × n 2 × . . . × n p possible outcomes for the p experiments. Proof. p = 2: the first task can be done in n 1 ways, for each of these ways there are n 2 choices for the second task. ( 1 × n 2 ) | {z } 1 + ( 1 × n 2 ) | {z } 2 + · · · |{z} ... + ( 1 × n 2 ) | {z } n 1 = n 1 × n 2 Assume true for p , i.e. n 1 × n 2 × · · · × n p ways to do the job. For p + 1, the ( p + 1 ) th task can be done in n p + 1 ways, and for each one of these ways we can complete the job by performing the remaining p tasks. Thus, ( 1 × ( n 1 × n 2 × · · · × n k )) + · · · + ( 1 × ( n 1 × n 2 × · · · × n p )) | {z } n p + 1 times = n 1 × n 2 × · · · × n p × n p + 1 7/33

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Examples: i. An eight-bit binary word is a sequence of eight digits, of which each may be either a zero or a one. There are 2 × 2 × 2 × 2 × 2
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