ch11applicationsofbinomialprobabilities.studentview

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Chapter 11 : Applications of Binomial Probabilities For repeated, independent, experiments with just two outcomes, success and failure, where P (success) = p and P (failure) = q = 1-p , the probability of gaining x successes out of n trials is ( 29 n x q p x X P x n x n x ,..... , 1 , 0 , ) ( = = = -
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The P(X=x) is called a binomial probability, and the whole list of P(x) for x=0,1,2,….,n is called a binomial distribution. We use the symbols X ~ Bin (n, p) to stand for the sentence, “X is distributed like a binomial model with number of trials n and probability of success per trial p”
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The values of the binomial probabilities can be computed directly. They are also tabulated for certain selected values of n and p in Table 2 which is an extract from a much larger book for binomial distributions. A: Straightforward Applications Example 1(a) : A plays B in the final of a badminton tournament and each performs independently from game to game. Whoever wins two games out of three first gets the gold medal. In each game, A’s chance of winning is 0.45 while B’s chance of winning is 0.55. What is the probability that A gets the medal?
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Solution : Assume that they play all the three games, whether or not the match ends before the 3 rd game. Let X = the number of games A wins in these three Then X ~ Bin( n = 3, p = 0.45) From Table 2, one finds, for n=3 and p=0.45, P(2)=0.3341 and P(3)=0.0911 Therefore, P(A gets gold medal) =P(X≥2) =P(X=2)+P(X=3)= 0.3342 + 0.0911 =0.4252
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Example 1(b) A plays B in the final of a badminton tournament. Who ever wins three games first gets the cup. In each game, A’s chance of winning is 0.65. Both perform independently from game to game. What is the probability that A gets the cup?
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7648 . 0 1160 . 0 3124 . 0 3364 . 0 = + + = Recommended Method : X = the number of times A wins X ~ Bin ( n=5, p=0.65) P(A gets the cup)=P(X=3) + P(X=4) + P(X=5) look up Binomial tables This method is recommended by Prof Chiu and Gabrielle!
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( 29 ( 29 ( 29 7648 . 0 ] 735 . 0 05 . 1 1 [ ) 65 . 0 ( ) 65 . 0 ( ) 35 . 0 ( ) 65 . 0 ( 6 ) 65 . 0 )( 35 . 0 ( ) 65 . 0 ( 3 ) 65 . 0 ( ) 65 . 0 ( ) 65 . 0 ( ) 35 . 0 ( ) 65 . 0 ( ) 65 . 0 )( 35 . 0 ( ) 65 . 0 ( ) 65 . 0 ( ) 65 . 0 ( 3 2 2 2 2 2 2 4 2 2 3 2 2 2 2 = + + = + + = + + = 1. Let’s look at Example 1(b) in more detail : 2. The way of working out the answer on this page is a method that many students like to do. 3. This method works BUT is NOT recommended by Gabrielle or Prof Chiu because it is not efficient (It’s correct but NOT fast!) 1. Who ever wins 3 games first! Possibilities: AA | A A=3 B=0 BAA | A ABA | A A=3 B=1 AAB | A BBAA | A BABA | A BAAB | A A=3 B=2 ABBA | A ABAB | A AABB | A Method B : P(A gets the cup)
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Recommended Method : X = the number of times A wins X ~ Bin ( n=5, p=0.65) P(A gets the cup)=P(X=3) + P(X=4) + P(X=5) look up Binomial tables This method is recommended and is MUCH FASTER than the other way Let’s look at a much BETTER (Prof Chiu’s) method: Who ever wins 3 games first! Possibilities:
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