Answer: d).0264
Step-by-step explanation:
Given : It is known that screws produced by a certain company will b defective with probability .01 independently of each other.
The company sells the screws in packages of 25 and offers a money back guarantee that at most 1 of the 25 is defective.
Let x be the binomial variable that represents the defective screws having parameter p= 0.01 and n= 25
For using Poisson approximation for binomial distribution
Mean = [tex]\lambda=np=25\times0.01=0.25[/tex]
Poisson distribution formula : [tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
Now, the probability that the company must replace a package is
=Probability that package has more than 1 defectives
=[tex]P(x>1)=1-(P(x\leq1))\\\\=1-[P(x=0)+P(x=1)]\\\\=1-[e^{-0.25}\cdot\frac{0.25^0}{0!}+e^{-0.25}\cdot\frac{0.25^1}{1!}]\\\\=1-0.9736=0.0264[/tex]
Hence, the required probability is 0.0264
