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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

\[P(2~out~of~4)=\binom{4}{2}~(bad)^2~(good)^2\]
\[P(2~out~of~4)=6~(.20)^2~(.80)^2\]

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 ?

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?