ch10A-binomialdistn

ch10A-binomialdistn - Chapter 10A: PERMUTATIONS and...

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Unformatted text preview: Chapter 10A: PERMUTATIONS and COMBINATIONS Example 1(a): Example 1(a) Determine the different PERMUTATIONS of three of the first four letters of the alphabet a, b, c, d and list them all. Solution: PERMUTATIONS­Order is important. abc acb bac bca cab cba abd adb bad bda dab dba acd adc cad cda dac dca bcd bdc cbd cdb dbc dcb RULE: nP = r n! ( n −r )! 4! P= =4! =4 × ×2 × =24 3 1 4 3 There are ways ( 4 −3)! Example 1 (b) If we are NOT concerned with the order then in how many ways can the selection be made? how Solution: [abc acb bac bca cab cba] {abd adb bad bda dab dba} (acd adc cad cda dac dca) <bcd bdc cbd cdb dbc dcb> COMBINATIONS - If we are NOT concerned with the order then: RULE: n! nP =r n Cr = (n − r )!r! r! 4 C3 = Here, there are only ! 3 2× ( ) = 3!(44− 3)! = 4 ×× × ×1 1 = 4 32 4 3 The committees are: a b c, a b d, a c d, bcd Example 2 (i): In how many ways can 8 different books be arranged on a shelf? Solution: Is the order important? Yes. Just ask a librarian!!! Solution: n! nP = r (n − r )! 8! 8! = = 8! 8P = 8 (8 − 8)! 0! There are 8! = 40,320 ways of arranging 8 different books on a shelf. Example 2(ii): Find the number of ways in which 4 of 10 teams can be ranked 1st, 2nd, 3rd and 4th? Example 2(ii): Solution: 10 x 9 x 8 x 7 ___ ___ ___ ___ 1st 2nd 3rd 4th n! nP = r (n − r )! 10 P = 4 10! 10! 10 × 9 × 8 × 7 × 6 × .... ×1 = = = 10 × 9 × 8 × 7 (10 − 4)! 6! 6 × .... ×1 There are 5,040 ways of ranking teams 1st, 2nd, 3rd and 4th. Example 2(iii): The five finalists of a beauty contest are Miss Argentina, Miss Belgium, Miss China, Miss Denmark, and Miss Egypt. In how many ways can the judges choose the champion, the first runner­up and the second runner­up? Solution: Is the order important? YES!!! Miss Argentina, Miss Belgium, Miss China, Miss Denmark, Miss Egypt 5 X 4 X 3 ___ ___ ___ 1st 2nd 3rd n! nP = r (n − r )! 5! 5! = 5P = 3 (5 − 3)! 2! 5 × 4 × 3 × 2 ×1 = 2 ×1 = 5× 4× 3 = 60 Example 3: In how many ways can a committee of 4 be selected from a group of 15 people? Solution: Is the order important? Solution: Is the order important? NO! It ‘s just who is on the committee So this is a COMBINATION 15 C4 = () 15 4 15! 15 ×14 ×13 ×12 = = = 1365 4!(11)! 4 × 3 × 2 ×1 Example 4(i): Among 20 applicants for 5 positions advertised on a newspaper, 9 are college graduates. If the selection is random, what is the probability that the positions will be filled by 3 graduates and 2 non­ graduates? Example 4(ii): An auto parts store has 200 rebuilt starters in stock, of which 4 are HYPERGEOMETRIC HYPERGEOMETRIC DISTRIBUTION DISTRIBUTION Solution: Example 4(i) ( )( ) = 84 x55 = 0.29799 P( E ) = ( ) 15504 9 3 11 2 20 5 Solution: Example 4(ii) ( )( ) = 6 x196 = 0.0009 P( E ) = ( ) 1,313,400 4 2 196 1 200 3 SUMMARY: SUMMARY: HYPERGEOMETRIC DISTRIBUTION HYPERGEOMETRIC Let X = the number of x “successes” in “n” trials then X ~ Hypergeom (N, R, n, x) where n = the number of trials N = the number of elements in the population R = the number of elements labeled “success” Note: This is the only distribution we study that is “WITHOUT REPLACEMENT” ( )( ) P( X = x) = () R x N −R n− x N n for 0 ≤ x ≤ R ...
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