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 one year 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)
 one year 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)

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1this might be better suited to a binomial setup

samnatha
 one year ago
Best ResponseYou've already chosen the best response.0thats what i did for part i and i got that i'm just not sure what to do for the other part

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[P(2~out~of~4)=\binom{4}{2}~(bad)^2~(good)^2\] \[P(2~out~of~4)=6~(.20)^2~(.80)^2\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the second part corresponds to an OR setup P(0) or P(1) or P(2)

samnatha
 one year ago
Best ResponseYou've already chosen the best response.0how do you do the second part i have done the first part ..i just need help with the second part

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[P(0,1,2~out~of~ 4)=6[(.2)^0 (.8)^4+(.2)^1 (.8)^3+(.2)^2 (.8)^2]\]

samnatha
 one year ago
Best ResponseYou've already chosen the best response.0where did u get the six from ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1hmm, 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

samnatha
 one year ago
Best ResponseYou've already chosen the best response.0ok so i got 0.9728 for my final answer ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1that does seem about right. can you see any other errors in my remembering of this?

samnatha
 one year ago
Best ResponseYou've already chosen the best response.0nope 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 :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so 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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