Page 163 - Distribusi peluang binomial

Contoh Masalah

Jawab P(X < 2) = P(X = 0) + P(X = 1)

Cara 2

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

= 1 – (P(X = 0) + P(X = 1))

= 1 – (b(0; 5; 90%) + b(1; 5; 90%))

= 1 – ( C (90%) . (10%) 5-0 + C (90%) . (10%) )
1
0
5-1
5 0
5 1
= 1 – ((1). (1).(0,00001) + (5). (0, 9).(0,0001))

= 1 – (0,00001 + 0,00045)

= 1 – 0,00046

= 0,99954

 0,9995