Solving Poisson Distribution: Part IV - Tank of Water (10^5 cm3)

[somecelxis]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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!

 

[somecelxis]

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!

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 probability 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 .

 

[haruspex]

Do you know that your answers to a, b, and c are correct? If not, please post your answers.
For (d), what are all the possible distributions of the 5 between the 6 test tubes?

 

[Ray Vickson]

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 probability 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 .

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.)

 

[somecelxis]

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.)

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?

 

[haruspex]

why the total volume is small compared with the overall tank volume, we can assume independence without making much of an error.??

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.

so my working would be (e^-0.36)x 0.36 = 0.2511

What are you calculating there, and how? Please show all steps.

 

[somecelxis]

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.

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

 

[Orodruin]

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

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?

 

[somecelxis]

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?

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?

 

[Orodruin]

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?

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?

 

[haruspex]

this is based on what the homework helper form physiscs forum have said .

If you are referring to my question:

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

please ignore that and follow Orodruin's hint.