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help please !! if 20% of the balls produced by the machine are defective find the probability that out of the 4 balls selected at random (i) 2 are defective (ii) not more than 2 are defective i have done part (i) i got 0.1536 but i need help with part (ii)

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

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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 :)

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