Respuesta :
The experiment has a binomial distribution just because the product could be or not defective. In this case you have n=26 the number of product that the company will check and you also know that the probability to get a defective one is 3.6%. Now we will proceed as follows, to find the probability of getting 2 or fewer defective products is exactly the same that
[tex]\begin{gathered} P(X\leq2)=P(X=0)+P(X=1)+P(X=2)\text{ } \\ \text{ where }X\text{ is a random variable that measures the number of defective products} \end{gathered}[/tex]We also should notice that
[tex]P(X=k)=(n;k)p^k(1-p)^{(n-k)};\text{ where }(n;k)=\frac{n!}{k!(n-k)!}[/tex]Now,
[tex]P(X=0)=(26;0)(0.036)^0(1-0.036)^{26-0}=(1-0.036)^{26}\approx0.3855[/tex][tex]P(X=1)=(26;1)(0.036)(1-0.036)^{26-1}=26(0.036)(1-0.036)^{25}\approx0.3743[/tex][tex]P(X=2)=(26;2)(0.036)²(1-0.036)^{26-2}=325(0.001296)(1-0.036)^{24}\approx0.1747[/tex]Then the probability that the company takes 2 or fewer defective products in this batch is
[tex]P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=0.3855+0.3743+0.1747=0.9345[/tex]Now, to get the probability that the company find 4 or more defective products we proceed as follows. Since
[tex]P(X\ge4)+P(X<4)=1,\text{ }[/tex]then
[tex]P(X\ge4)=1-P(X<4)=1-(\text{ }P(X=0)+P(X=1)+P(X=2)+P(X=3)).[/tex]Notice that we only need to find one of these probabilities to get the answer.
[tex]P(X=3)=(26;3)(0.036)³(1-0.036)^{26-3}=\frac{23!(24)(25)(26)}{3!23!}(0.036)³(1-0.036)^{23}=0.05219[/tex]Then the probability to get more than 4 defective products is
[tex]\begin{gathered} P(X\ge4)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))= \\ =1-(0.9345+0.05219)=0.01331 \\ \\ P(X\ge4)=0.01331 \end{gathered}[/tex]So if the company finds 5 defective products in this batch the company should stop the production because the probability to get more than 4 defective products is around 1.3%. It means that if you get 5 defective products even with this small probability then we could say that is really probable to get more defective products that those who were expected.
