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this might be better suited to a binomial setup
thats what i did for part i and i got that i'm just not sure what to do for the other part
the second part corresponds to an OR setup P(0) or P(1) or P(2)
how do you do the second part i have done the first part ..i just need help with the second part
\[P(0,1,2~out~of~ 4)=6[(.2)^0 (.8)^4+(.2)^1 (.8)^3+(.2)^2 (.8)^2]\]
where did u get the six from ?
hmm, i do seem to have been a little amiss with that .... apparently we are not choosing 2 out of 4 each time :/ (4 0) = 1 (4 1) = 4 (4 2) = 6
ok so i got 0.9728 for my final answer ?
is that right ?
that does seem about right. can you see any other errors in my remembering of this?
nope although i wouldn't have added it up to 6 i would have done each sum seperatly and then added them but thank you for your help :)
so the liklihood that no more than 2 are defective out of 4 is about 98%, sounds like a winner to me ... good luck :)