El diseño de un sistema requiere instalación de 2 componentes iguales. El sistema completo funcionará si al menos 1 de los componentes funcionan,mientras que un sistema alternativo requiere cuatro de sus componentes funcionará si al menos 2 de sus componentes están funcionando.
La probabilidad de que un componente funcione de manera individual es de 0.9 y cada uno funciona de manera independiente. ¿cuál de los 2 diseños tienen mayor probabilidad de funcionar?

Usando la distribución binomial, se encuentra que el diseño de 4 componentes tiene mayor probabilidad de funcionar.

Para cada componente, solo hay dos resultados posibles. O están funcionando o no estan. La probabilidad de que un compoente estea funcionando es independiente de cualquier otro componente, lo que significa que se utiliza la distribución binomial para resolver esta cuestión.

Distribuición binomial:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Los parámetros son:

En este problema:

Diseño de 2 componentes:

La probabilidad de funcionar es [tex]P(X \geq 1)[/tex], que es dada por:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

En que:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{2,0}.(0.9)^{0}.(0.1)^{2} = 0.01[/tex]

Entonces:

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.01 = 0.99[/tex]

Diseño de 4 componentes:

La probabilidad de funcionar es [tex]P(X \geq 2)[/tex], que es dada por:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

En que:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{4,0}.(0.9)^{0}.(0.1)^{4} = 0.0001[/tex]

[tex]P(X = 1) = C_{4,1}.(0.9)^{1}.(0.1)^{3} = 0.0036[/tex]

Entonces:

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0001 + 0.0036 = 0.0037[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0037 = 0.9963[/tex]

[tex]0.9963 > 0.99[/tex], o sea, el diseño de 4 componentes tiene mayor probabilidad de funcionar.

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