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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)
 11 months ago
 11 months ago
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)
 11 months ago
 11 months ago

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amistre64Best ResponseYou've already chosen the best response.1
this might be better suited to a binomial setup
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
thats what i did for part i and i got that i'm just not sure what to do for the other part
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
\[P(2~out~of~4)=\binom{4}{2}~(bad)^2~(good)^2\] \[P(2~out~of~4)=6~(.20)^2~(.80)^2\]
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
the second part corresponds to an OR setup P(0) or P(1) or P(2)
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
how do you do the second part i have done the first part ..i just need help with the second part
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
\[P(0,1,2~out~of~ 4)=6[(.2)^0 (.8)^4+(.2)^1 (.8)^3+(.2)^2 (.8)^2]\]
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
where did u get the six from ?
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
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
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
ok so i got 0.9728 for my final answer ?
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
that does seem about right. can you see any other errors in my remembering of this?
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
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 :)
 11 months ago

amistre64Best ResponseYou've already chosen the best response.1
so the liklihood that no more than 2 are defective out of 4 is about 98%, sounds like a winner to me ... good luck :)
 11 months ago
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