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

- anonymous

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- anonymous

@nincompoop

- anonymous

@IrishBoy123

- anonymous

@wio

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## More answers

- anonymous

yesss.. actually am learning poisson distribution

- anonymous

and ?

- 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

how would you calculate them ?

- alekos

P(2 or 3 or 4 defects) = P(2 defects) + P(3 defects) + P(4 defects)

- anonymous

help me with the numbers

- alekos

well you already know that P(1 defect) = 1/10000 so from there it is just simple arithmetic

- anonymous

yess

- alekos

use your calculator in exponential or scientific mode

- anonymous

are you sure its simple arithmetic ? hve u calculated ?

- alekos

yes, I sure have

- anonymous

what ans uve got ?

- alekos

Approximately
\[\frac{ 1 }{ 10^{8} }\]

- anonymous

its wrong.. the answer should be 0.471
its not simple arithmetic

- anonymous

aniway thanks

- alekos

can you send me a photo of the actual question?

- anonymous

it is the actual question

- alekos

That figure is not possible based on the information you have given

- IrishBoy123

i can help you but not whilst others are as that is counter productive for all concerned
tag me later

- anonymous

do you know poisson distribution ?

- alekos

go for it IrishBoy

- IrishBoy123

yes
and the answer is 0.471004095.......

- IrishBoy123

thx @alekos :p

- IrishBoy123

start with the formula and work out \( \lambda \). have you done that? expected outcome given failure rate of 1/10000

- anonymous

yes

- 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

\( \lambda\) 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

thank you

- 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

yeah.. got all

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