anonymous
  • anonymous
pleeeeeeeeeeeeaaaaaaaase helllllp me.......... a certain machine make matches. one match in 10000 on average is defective. using a suitable approximation, find the probability that a random sample of 45000 matches will include 2, 3 or 4 defective matches
Mathematics
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@nincompoop
anonymous
  • anonymous
@IrishBoy123
anonymous
  • anonymous
@wio

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anonymous
  • anonymous
yesss.. actually am learning poisson distribution
anonymous
  • anonymous
and ?
alekos
  • alekos
P(2 defects) = P(1 defect) x P(1 defect) P(3 defects) = P(2 defects) x P(1 defect) P(4 defects) = P(3 defects) x P( 1 defect)
anonymous
  • anonymous
how would you calculate them ?
alekos
  • alekos
P(2 or 3 or 4 defects) = P(2 defects) + P(3 defects) + P(4 defects)
anonymous
  • anonymous
help me with the numbers
alekos
  • alekos
well you already know that P(1 defect) = 1/10000 so from there it is just simple arithmetic
anonymous
  • anonymous
yess
alekos
  • alekos
use your calculator in exponential or scientific mode
anonymous
  • anonymous
are you sure its simple arithmetic ? hve u calculated ?
alekos
  • alekos
yes, I sure have
anonymous
  • anonymous
what ans uve got ?
alekos
  • alekos
Approximately \[\frac{ 1 }{ 10^{8} }\]
anonymous
  • anonymous
its wrong.. the answer should be 0.471 its not simple arithmetic
anonymous
  • anonymous
aniway thanks
alekos
  • alekos
can you send me a photo of the actual question?
anonymous
  • anonymous
it is the actual question
alekos
  • alekos
That figure is not possible based on the information you have given
IrishBoy123
  • IrishBoy123
i can help you but not whilst others are as that is counter productive for all concerned tag me later
anonymous
  • anonymous
do you know poisson distribution ?
alekos
  • alekos
go for it IrishBoy
IrishBoy123
  • IrishBoy123
yes and the answer is 0.471004095.......
IrishBoy123
  • IrishBoy123
thx @alekos :p
IrishBoy123
  • IrishBoy123
start with the formula and work out \( \lambda \). have you done that? expected outcome given failure rate of 1/10000
anonymous
  • anonymous
yes
IrishBoy123
  • IrishBoy123
then just plug in as per @alekos suggestion ie P(2 or 3 or 4 defects) = P(2 defects) + P(3 defects) + P(4 defects) but what is your \( \lambda\)? that's all that's missing
IrishBoy123
  • IrishBoy123
\( \lambda\) is how many you would expect to be defective in a lot of 45,000 given failure rate 1/10,000 then \( P(X=2) = \frac{e^{- \lambda } \lambda^2}{2!}\) do same for 3 and 4 and add and you're done
anonymous
  • anonymous
thank you
IrishBoy123
  • IrishBoy123
i have to go but here are the numbers you should get 2 0.11247859 3 0.168717885 4 0.189807621 \(\Sigma\) 0.471004095
anonymous
  • anonymous
yeah.. got all

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