A factory produces light bulbs with a defect rate of 2%. If 100 light bulbs are selected at random, what is the probability of finding exactly 3 defective light bulbs?

1. Identificar los valores de \( n \), \( k \), y \( p \):

n = 100

k = 3

p = 0.02

2. Calcular el coeficiente binomial:

\binom{100}{3} = \frac{100!}{3!(100-3)!} = \frac{100!}{3! \cdot 97!} = \frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1} = 161700

3. Calcular \( p^k \) y \( (1-p)^{n-k} \):

p^k = (0.02)^3 = 0.000008

(1-p)^{n-k}=(0.98)^{97}\approx0.1409

4. Sustituir en la fórmula de la distribución binomial y calcular la probabilidad:

P(X=3)=161700\cdot0.000008\cdot0.1409\approx0.1823

Respuesta:

P(X=3)\approx0.1823