slides6 - Lecture Stat 302 Introduction to Probability -...

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Lecture Stat 302 Introduction to Probability - Slides 6 AD Jan. 2010 AD () Jan. 2010 1 / 12
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Independence We say that the events f E i g n i = 1 are independent if P ( \ n i = 1 E i ) = n i = 1 P ( E i ) . and if for i 1 , i 2 , ..., i r where i j 6 = i k for j , k 2 f 1 , ..., r g and any r 2 f 1 , ..., n g we have P \ ± E i j ² j = 1 ,..., r ³ = r j = 1 P ´ E i j µ . Example : We say that three events E , F , G are independent if P ( E \ F \ G ) = P ( E ) P ( F ) P ( G ) and if P ( E \ F ) = P ( E ) P ( F ) , P ( E \ G ) = P ( E ) P ( G ) , P ( F \ G ) = P ( F ) P ( G ) . AD () Jan. 2010 2 / 12
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Example: Independent Trials proba p it is a success and with proba 1 p this is a failure. What is n trials 2) exactly k n trials 3) all trials results in successes? Answer: 1) Let E i event of a failure at trial i then proba of no success is P ( E 1 \ E 2 \±±±\ E n ) = n i = 1 P ( E i ) = ( 1 p ) n so proba of at least one success is 1 ( 1 p ) n . 2) Proba of exactly k successes is n k ± p k ( 1 p ) n k . 3) Proba of all successes is P ( \ i = 1 E c i ) = lim n ! P ( \ n i = 1 E c i ) = lim n ! p n = 0 for p < 1. AD () Jan. 2010 3 / 12
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Example: The problem of points proba. p it is a success and with proba 1 p this is a failure. What is the proba of getting at least n successes before getting m failures? Answer: For at least n successes to occur before m failures, we need at least n n + m 1 trials. Hence, we have P ( at least n successes ) = n + m 1 k = n P ( k successes exactly ) = n + m 1 k = n n + m 1 k ± p k ( 1 p ) m + n 1 k AD () Jan. 2010 4 / 12
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Example: Dice. .. again
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slides6 - Lecture Stat 302 Introduction to Probability -...

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