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Lecture One

Lecture One. Events & Probability. Example . Two dice are thrown once. Let A , and B be two events defined by: A = the first die shows the number 1 . B = the sum of two numbers appearing is less than 6 . Find : P(A) , P(B) , P(A B) , P(A B C ) , P(A∩B) Solution :

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Lecture One

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  1. Lecture One Events & Probability

  2. Example . Two dice are thrown once. Let A , and B be two events defined by: • A = the first die shows the number 1. • B = the sum of two numbers appearing is less than 6. • Find : P(A) , P(B) , P(A B) , P(A BC) , P(A∩B) • Solution: • A= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} • A∩B= {(1,1), (1,2), (1,3), (1,4)} • P(A BC) = P(A) + P(BC) - P(A∩BC) • P(BC) = 1- P(B) • P(A∩BC) = P(A) - P(A∩B)

  3. Conditional Probability • Example .There three urns A, B & C. Urn A contains six balls {three red, two black, one white}. Urn B contains four balls {one red, one black, two white}. Urn C contains twelve balls {three red, five black, four white}. We select one ball from these urns "randomly", and then select two balls after that we notice one of them is white and the other is red. What is the probability these two balls were selected from urn A? • Solution: Under the principle of "equally likely", then: • Pr(A) = Pr(B) = Pr(C) = 1/3

  4. E = Two balls are selected regardless the urns {A, B or C}. • E1 = Two balls are selected from urn A. • E2 = Two balls are selected from urn B. • E3 = Two balls are selected from urn C.

  5. Problem • In a trial, the judge is 65% sure that Susan has committed a crime. Julie and Robert are two witnesses who know whether Susan is innocent or guilty. However, Robert is Susan’s friend and will lie with probability 0.25 if Susan is guilty. He will tell the truth if she is innocent. Julie is Susan’s enemy and will lie with probability 0.30 if Susan is innocent. She will tell the truth if Susan is guilty. What is the probability that Susan is guilty if Robert and Julie give conflicting testimony?

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