# Poisson distribution

1. Aug 21, 2014

### somecelxis

1. The problem statement, all variables and given/known data

i am having problem with part iv ) . the ans is 0.04519 . can anyone tell me how to do this ? i have solved part i , ii and iii ..p/s line 1:
A tank contain 10^5 cm3 of water
2. Relevant equations

3. The attempt at a solution

#### Attached Files:

File size:
68.2 KB
Views:
73
• ###### IMG_20140820_172031[1].jpg
File size:
31.4 KB
Views:
70
2. Aug 21, 2014

### Ray Vickson

You seem to make a habit of violating PF standards: you keep posting thumbnails, when--officially--you are supposed to type things out. I cannot read your thumbnails on some media, and on the computer I am using now the part you want to know about (part (iv)) does not even show up in the attached picture!

3. Aug 21, 2014

### somecelxis

A water tank contains 10^5 cm3 of water that is free from bacteria. A number of 1800 bacteria were added to water and water is stirred to disrtibute the bactria randomly. Then 6 test tube were filled with 20cm3 of this water. Findthe probablity of
a) a test tube contain 3 bactria
b) A test tube contain at least 3 bacteria
c) exactly 4 out of 6 test tube with each contain 3 bacteria

d) 6 test tube conatin a total of 5 bacteria

i got stucked at part d .

4. Aug 22, 2014

### haruspex

For (d), what are all the possible distributions of the 5 between the 6 test tubes?

5. Aug 22, 2014

### Ray Vickson

The bacteria amount in 6 test tubes is the sum of 6 independent, identically-distributed Poisson random variables. Do you know what that distribution is? (Of course, in reality the 6 test-tube bacteria contents are not really independent, but since the total volume is small compared with the overall tank volume, we can assume independence without making much of an error.)

Last edited: Aug 22, 2014
6. Aug 22, 2014

### somecelxis

why the total volume is small compared with the overall tank volume, we can assume independence without making much of an error.??
so my working would be (e^-0.36)x 0.36 = 0.2511
step 2 : 6c5 ( (0.2511)^5 )x ( (1-0.2511)^1 )= 4.49x10^-3
do you mean we can assume for 6 test tube , each conatain only exactly one bacteria?

7. Aug 22, 2014

### haruspex

That wasn't quite complete. There are two facts which, together, mean you can treat the samples as independent:
1. The sample volumes are tiny compared to the tank volume
2. The number of bacteria sampled, in total, is small compared with the total in the tank.
As a result, taking one sample does not much change the circumstances for the next sample.
What are you calculating there, and how? Please show all steps.

8. Aug 22, 2014

### somecelxis

6c5 ( (0.2511)^5 )x ( (1-0.2511)^1 )= 4.49x10^-3
i assume each tube contain only 1 bacteria.

9. Aug 22, 2014

### Orodruin

Staff Emeritus
By making this assumption you are effectively saying that the only way of catching 5 bacteria is to catch 1 in each of 5 samples and getting the last sample empty. What about the possibility of catching all bacteria in one sample? What about that of catching 3 in one, 2 in one, 1 in one, and none in 3?

Does it matter which of the samples the bacteria are caught in?

10. Aug 22, 2014

### somecelxis

this is based on what the homework helper form physiscs forum have said . If i consider catching 3 in one, 2 in one, 1 in one, and none in 3?

then i should also consider catching 3 in two, catching 4 in two , catching 5 in two , there would be too many combinations . any other simpler way look at tis question?

11. Aug 22, 2014

### Orodruin

Staff Emeritus
Yes, in fact this I would say is the entire point. It has already been indicated above (see the post by Ray Vickson) but you can think of it in different ways.

If you take 6 samples of 20 cm3 each, how much water have you taken out and how many bacteria would you expect to be in that water?

12. Aug 22, 2014

### haruspex

If you are referring to my question: