## anonymous one year ago 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

1. anonymous

@nincompoop

2. anonymous

@IrishBoy123

3. anonymous

@wio

4. anonymous

yesss.. actually am learning poisson distribution

5. anonymous

and ?

6. 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)

7. anonymous

how would you calculate them ?

8. alekos

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

9. anonymous

help me with the numbers

10. alekos

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

11. anonymous

yess

12. alekos

use your calculator in exponential or scientific mode

13. anonymous

are you sure its simple arithmetic ? hve u calculated ?

14. alekos

yes, I sure have

15. anonymous

what ans uve got ?

16. alekos

Approximately $\frac{ 1 }{ 10^{8} }$

17. anonymous

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

18. anonymous

aniway thanks

19. alekos

can you send me a photo of the actual question?

20. anonymous

it is the actual question

21. alekos

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

22. IrishBoy123

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

23. anonymous

do you know poisson distribution ?

24. alekos

go for it IrishBoy

25. IrishBoy123

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

26. IrishBoy123

thx @alekos :p

27. IrishBoy123

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

28. anonymous

yes

29. 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

30. 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

31. anonymous

thank you

32. 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

33. anonymous

yeah.. got all