Page 162 - Distribusi peluang binomial
P. 162

Contoh masalah







 Jawaban untuk c yaitu “paling sedikit 2 kalkulator baik”







 Cara 1








 P(X  2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)




 = b(2; 5; 90%) + b(3; 5; 90%) + b(4; 5; 90%) +




    b(5; 5; 90%)





 =  C  (90%) . (10%) 5-2  +  C  (90%) . (10%) 5-3     +  C  (90%) . (10%)                                               5-4      +
                                                                                              4
                   3
 2
 5 2
 5 3
                                                            5 4
 C  (90%) . (10%)
 5 5  5  5-5

 = (1/2)5(4). (0,81).(0,001) + (1/6)5(4)(3). (0,729).(0,01) +





   (5). (0,6561).(0,1) + (1). (0,59049).(1)




 = 0,0081 + 0,0729 + 0,32805 + 0,59049




 = 0,99954




  0,9995
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