The probability that one 3 foot section of wire is defective is 0.002. If someone has 450 feet of wire then what is the probability they will have 3 or more 3 feet sections that are defective?
don't really know where to start, step by step would be awesome. i have a test tomrorow.
don't really know where to start, step by step would be awesome. i have a test tomrorow.
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Jack -
This fits a binomial, but with the small p value of .002, the Poisson will be easier to calculate:
Let λ = np = 150(02) = 0.3 , [Note: the wire has 150 sections that equal 3 feet]
For Poisson, p(x) = (λ^x)(e^- λ) / x!
Now, to find the probability 3 or more 3 feet sections that are defective:
P(x >=3) = 1 - P(0) - P(1) - P(2) , are you with me? It is easier to solve this problem by looking at the complement.
P(0) = (0.3^0)(e^- 0.3) / 0! = 0.7408
P(1) = 0.2222
P(2) = 0.0333
P(x >=3) = 1 - P(0) - P(1) - P(2) = 1 - 0.7408 - 0.2222 - 0.0333 = 0.0037
Hope that helped
P.S. - if you would have used Binomial, the answer would be 0.0035
This fits a binomial, but with the small p value of .002, the Poisson will be easier to calculate:
Let λ = np = 150(02) = 0.3 , [Note: the wire has 150 sections that equal 3 feet]
For Poisson, p(x) = (λ^x)(e^- λ) / x!
Now, to find the probability 3 or more 3 feet sections that are defective:
P(x >=3) = 1 - P(0) - P(1) - P(2) , are you with me? It is easier to solve this problem by looking at the complement.
P(0) = (0.3^0)(e^- 0.3) / 0! = 0.7408
P(1) = 0.2222
P(2) = 0.0333
P(x >=3) = 1 - P(0) - P(1) - P(2) = 1 - 0.7408 - 0.2222 - 0.0333 = 0.0037
Hope that helped
P.S. - if you would have used Binomial, the answer would be 0.0035