Respuesta :
Answer:
A. P=0.4013
B. P=0.9884
C. P=0.3222
Step-by-step explanation:
We can model this with a binomial random variable.
The sample size is n=10.
The probability is p=0.05 for scrap and p=0.05+0.15=0.20 for degraded or scrap.
The probability of having k scrap gears in the sample is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{10}{k} 0.05^{k} 0.95^{10-k}\\\\\\[/tex]
The probability that one or more is scrap can be calculated as 100% less the probability that no one is scrap:
[tex]P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x\geq1)=1-0.5987=0.4013[/tex]
The probability that 8 or more are not scrap is equal to the probability of having 2 or less that are scrap:
[tex]P(x\leq2)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.05*0.6302=0.3151\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.0025*0.6634=0.0746\\\\\\P(x\leq2)=0.5987+0.3151+0.0746=0.9884[/tex]
The probability that more than two are degraded or scrap (p=0.2) is calculated as:
[tex]P(x>2)=1-P(x\leq2)=1-(P(x=0)+P(x=1)+P(x=2))\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.1074=0.1074\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.2*0.1342=0.2684\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.04*0.1678=0.3020\\\\\\P(x>2)=1-(0.1074+0.2684+0.3020)=1-0.6778=0.3222[/tex]
