Question

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

Answer 1

@nincompoop

Answer 2

@IrishBoy123

Answer 3

@wio

Answer 4

yesss.. actually am learning poisson distribution

Answer 5

and ?

Answer 6

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)

Answer 7

how would you calculate them ?

Answer 8

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

Answer 9

help me with the numbers

Answer 10

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

Answer 11

yess

Answer 12

use your calculator in exponential or scientific mode

Answer 13

are you sure its simple arithmetic ? hve u calculated ?

Answer 14

yes, I sure have

Answer 15

what ans uve got ?

Answer 16

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

Answer 17

its wrong.. the answer should be 0.471

its not simple arithmetic

Answer 18

aniway thanks

Answer 19

can you send me a photo of the actual question?

Answer 20

it is the actual question

Answer 21

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

Answer 22

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

tag me later

Answer 23

do you know poisson distribution ?

Answer 24

go for it IrishBoy

Answer 25

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

Answer 26

thx @alekos :p

Answer 27

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

Answer 28

yes

Answer 29

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

Answer 30

\( \lambda\) 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

Answer 31

thank you

Answer 32

i have to go but here are the numbers you should get

2 0.11247859
3 0.168717885
4 0.189807621
\(\Sigma\) 0.471004095

Answer 33

yeah.. got all