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 .
So there are an average of 1800/10000= 18/100= 9/50 "bacteria per cm".
(9/50)(20)= 18/5 bacteria per test tube.Then 6 test tube were filled with 20cm3 of this water. Findthe probablity of
These two problems can be done using the Poisson distribution with $\displaystyle \lambda= 18/5$. (a) is P(3) and (b) is 1- (P(0)+ P(1)+ P(2)).a) a test tube contain 3 bactria
b) A test tube contain at least 3 bacteria
This requires, as romsek said, the binomial distribution with p= the probability you got in (a)c) exactly 4 out of 6 test tube with each contain 3 bacteria
d) 6 test tube conatin a total of 5 bacteria [/quote]
How many different ways can 5 bacteria be in 6 specified test tubes? They could all 5 be in one test tube, or 5 in one test tube and one in another, or 3 in one test tube and 2 in another, or 3 in one test tube and 1 in each of two other test tubes, etc.
i got stucked at part d .[/QUOTE]
i got stucked at part d .[/QUOTE][/QUOTE]
d) 6 test tube conatin a total of 5 bacteria [/quote]
How many different ways can 5 bacteria be in 6 specified test tubes? They could all 5 be in one test tube, or 5 in one test tube and one in another, or 3 in one test tube and 2 in another, or 3 in one test tube and 1 in each of two other test tubes, etc.
for part d , can you show me all the possibility of having 5 bacteria in 6 test tube? any other easy method to get the ans for this? it seems quite complicated if i do in this way..