Supongamos que el 20% de los objetos fabricados por una empresa son defectuosos. Escogemos 4 objetos al azar. Hallar la probabilidad de que (a) 2 sean defectuosos, (b) 3 sean defectuosos, (c) ninguno sea defectuoso.

Responder:

0,1536

0.0256

0.4096

Explicación paso a paso:

Dado que :

p = 0,2

Número de ensayos, n = 4

1 - p = 1 - 0,2 = 0,8

Probabilidad de que 2 sean defectuosos;

Usando la fórmula de distribución binomial:

P (x = 2)

P (x = x) = nCx * p ^ x * (1 - p) ^ (n - x)

P (x = 2) = 4C2 * 0.2 ^ 2 * 0.8 ^ 2

P (x = 2) = 6 * 0,04 * 0,64

P (x = 2) = 0,1536

3 están defectuosos:

P (x = 3) = 4C3 * 0,2 ^ 3 * 0,8 ^ 1

P (x = 3) = 4 * 0,008 * 0,8

P (x = 3) = 0,0256

Probabilidad de que ninguno sea defectuoso:

P (x = 0) = 4C0 * 0.2 ^ 0 * 0.8 ^ 4

P (x = 0) = 1 * 1 * 0,4096

P (x = 0) = 0,4096

the answer is (-4,-4)

hope this ! : d

answer: what is the conflict?

step-by-step explanation:

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