this is a binomial probability ... defective or not defective
probability (d) = .08 , probability (n) = .92 ... (n + d)^12
expand the binomial and substitute the probabilities in the desired terms
... n^12 + 12 n^11 d + 66 n^10 d^2 + ... + 66 n^2 d^10 + 12 n d^11 + d^12
less than 3 ... .92^12 + (12 * .92^11 * .08) + (66 * .92^10 * .08^2)
p(more than 2) = 1 - p(less than 3)
Suppose you work for the quality control department at Hewlett Packard. You know that ink jet cartridges are sometimes defective. The current defective rate is 8 cartridges per package of 100 (i.e. 0.08) What is the probability of sampling 12 cartridges and finding:
Less than 3 are defective:
More than 2 are defective:
At least 4 are defective:
Exactly 1 is defective:
2 or more, but less than 5 is defective:
If you could help me on at least 1 I would greatly appreciate that, or even point me where I need to go, I have no clue what to do here.
8 answers
I'm so sorry, but is there any way you can break down the whole (12 * .92^11 * .08) + (66 * .92^10 * .08^2) part for me. I'm just not understanding where all of these values came from. aside from 0.92 and 0.08
when you expand the binomial , the 12, 66, etc. are the coefficients of the terms
look up Pascal's Triangle for background
look up Pascal's Triangle for background
So if I'm understanding this right, for the probability of it being less than 3 my answer should be 0.93 right?
yes ... .9348
More than 2 would be 0.06?
Can you walk me through what I need to do for the last 3 please?
Can you walk me through what I need to do for the last 3 please?
at least 4 ... 1 - (first 4 terms)
Exactly 1 ... second term
2 or more, but less than 5 ... (first 5 terms) - (first 2 terms)
Exactly 1 ... second term
2 or more, but less than 5 ... (first 5 terms) - (first 2 terms)
Less than 3 are defective
---> none defective + one defective + 2 defective
= C(12,0)(.08)^0 (.92)^12 + C(12,1)(.08)^1 (.92)^11 + C(12,2)(.08)^2 (.92)^10
= .367666... + .38365... + .18348..
= appr .9348
More than 2 are defective
---> 3 defective + 4 def + ... + 12 def.
or 1 - (none def + 1 def + 2 def)
= 1 - (C(12,0)(.92^0)(.08^12) + C(12,1)(.92^1)(.08^11) + C(12,2)(.92^2)(.08^10) )
= ...
do you get the idea?
---> none defective + one defective + 2 defective
= C(12,0)(.08)^0 (.92)^12 + C(12,1)(.08)^1 (.92)^11 + C(12,2)(.08)^2 (.92)^10
= .367666... + .38365... + .18348..
= appr .9348
More than 2 are defective
---> 3 defective + 4 def + ... + 12 def.
or 1 - (none def + 1 def + 2 def)
= 1 - (C(12,0)(.92^0)(.08^12) + C(12,1)(.92^1)(.08^11) + C(12,2)(.92^2)(.08^10) )
= ...
do you get the idea?
3 answers