A combinatorial probability question

[tstl]

I've a small molecule ABCD made of building blocks A,B,C,D. The molecule can get cleaved at any 'bond' between the building block one bond at a time.
Researchers have compiled probabilities from dissociation studies of large number of molecules( made up of many other building blocks) and the dissociation probs are as follows say:
A:B = 0.1, B:C = 0.7, C:D = 0.3
So, when the bond between A-B breaks, I get a FRAGMENT PAIR with some prob (0.1) plus the molecule ABCD has some prob. 1-0.1 = 0.9 to remain undissociated.
However, as these probs are derived from a large dataset I've to normalize the probabilities so that, the sum of the probabilities of the exhaustive events becomes one for my case.
Hence,
the normalizing constant (k) would follow:
k * ( 0.1 + (1-0.1) + 0.7 + (1- 0.7) + 0.3 + ( 1- 0.3) ) = 1.0
or, k * 3 = 1.0
or, k = 1/3 = 0.33

So. the normalized probs of fragment pairs
p(A, BCD) = 0.1 * 0.33 =0.033
p(AB, CD) = 0.7 * 0.33 = 0.231
p(ABC, D) = 0.3 * 0.33 = 0.099

and the total prob. of ABCD to remain un-fragmented = (1-0.1) *0.33 + (1-0.7) * 0.33 + (1-0.3) *0.33 = 0.297 + 0.099 + 0.231 = 0.627

Am I doing it right?
Thanks

 

[PeroK]

I'm not sure I understand the data. Does that diagram come with the question?

 

[docnet]

The total probability that ABCD remains un-fragmented, would equivalent to computing $P((A:B)^c\cap (B:C)^c\cap (C:D)^c)$,using the fact that the events $A:B$, $B:C$, and $C:D$ are independent.

 

[docnet]

If you want to consider consider the probabilities of finding A, BCD, AB, CD, ABC, D, and ABCD in a mixture where all 7 types are mixed, then for the normalization constant, you'd have to add up the probabilities of finding each type of molecule and divide 1 by the sum.

 

[tstl]

Thanks docnet. Yes I want to consider the probabilities of finding the all 7 types. However I'm not sure what'll be the probability of each type say A and BCD when A:B breaks. Will it be (1/2) * 0.1 and (1/2) * 0.1 respectively (to be normalized later) because the Fragment pair has a probability of 0.1?

Secondly, can you show how to find the complement and intersection of the sets as you've mentioned to find out the prob. of ABCD remains un-fragmented

Thanks again

 

[docnet]

Thanks docnet. Yes I want to consider the probabilities of finding the all 7 types. However I'm not sure what'll be the probability of each type say A and BCD when A:B breaks. Will it be (1/2) * 0.1 and (1/2) * 0.1 respectively (to be normalized later) because the Fragment pair has a probability of 0.1?

Secondly, can you show how to find the complement and intersection of the sets as you've mentioned to find out the prob. of ABCD remains un-fragmented

Thanks again

I believe you are almost there, except no multiplying by (1/2), because A and BCD both have a (not normalized and relative) probability of .1 to be present in the mix. Then you would have to multiply the term P(ABCD) by (1/2) too, otherwise I believe you would be accounting for each block of ABCD twice.

Probability of the complement can be found using $P(A^c)=1-P(A)$. You will also need '$P(A \text{ and } B)=P(A)P(B)$ when $A$ and $B$ are independent'.

 

[tstl]

Thanks Docnet,
So here are the probabilities before normalization of the 7 species after one bond breaking:
P(A) = P(BCD) = 0.1
P(AB) = P(CD) = 0.7
P(ABC) = P(D) = 0.3
P(ABCD) = (1 - 0.1) * (1-0.7) * (1-0.3) = 0.9 * 0.3 * 0.7 = 0.189
Is this OK?

I was previously doing P(ABCD) = (1 - 0.1) + (1-0.7) + (1-0.3). Can you tell me why to multiply instead of taking the sum of probabilities?

 

[PeroK]

The scenario is this: we start with a large number of molecules of ABCD. After a certain time, some of the molecules have split into two molecules A + BCD; some have split into AB + CD; some have split into ABC + D; and some have not split and remain ABCD. Each of these processes has a certain probability:

There is a probability $p_0$ that the molecule stays as ABCD. Is $p_0 = 0.9$? Or is $p_0 = 0.1$? Or is $p_0$ something else? It's not clear from what you've written what all these numbers mean.

There are probabilities $p_1, p_2, p_3$ that the molecule splits into A + BCD, AB + CD, ABC + D. It's not at all clear what these probabilities are. They cannot be 0.1, 0.7 and 0.3, as these add up to 1.1. What do these numbers (0.1, 0.7, 0.3) mean?

I'm sorry that I can make no sense of what you are trying to do.

 

[docnet]

In probability theory, when there are independent events $A$ and $B$, and you want the probability of event $A$ and $B$ happening together, it's defined as $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B).$ Don't ask me why it's multiplication instead of addition, I don't know. It just works.

I think it's also a bit weird that $ABCD$ can only be broken once between two adjacent letters and we are considering a mixture that allows all 3 types of cleavings.

It's not at all clear what these probabilities are. They cannot be 0.1, 0.7 and 0.3, as these add up to 1.1. What do these numbers (0.1, 0.7, 0.3) mean?

I interpreted it as 'the reaction ABCD -> A, BCD has a .1 probability of occurring, meaning that after equilibrium, approximately $1/11$ of the pieces are A, and $1/11$ pf the pieces are BCD, and $9/11$ of the pieces are ABCD.'

 

[docnet]

What class is this for by the way? I'm asking because if we know the level of the class, we can perhaps help better by knowing what level of work is expected.

 

[PeroK]

I interpreted it as 'the reaction ABCD -> A, BCD has a .1 probability of occurring, meaning that after equilibrium, approximately $1/11$ of the pieces are A, and $1/11$ pf the pieces are BCD, and $9/11$ of the pieces are ABCD.'

What about the other reactions?

 

[PeroK]

Maybe the OP is expected to use these probabilities that give the equilibrium concentrations, to compute individual reaction constants then and create an ICE table to compute the final concentration of some given reaction that combines two or more of the individual reactions. It's hard to know.

There may be some confusion about whether the numbers given are probabilities of a reaction taking place or (relative) concentrations found in a mixture.

 

[docnet]

It seems most likely that OP is expected to compute the relative equilibrium concentrations (so they sum to 1 over the two sides) of the reaction
$$ABCD \rightarrow A+ BCD+AB+ CD+ABC+D.$$

There may be some confusion about whether the numbers given are probabilities of a reaction taking place or (relative) concentrations found in a mixture.

It seems like an introductory level science course, so it seems doubtful that it would be that heavy on probability theory. They probably are just the relative concentrations.

Edited grammar.

 

[PeroK]

It seems most likely that OP is expected to compute the relative equilibrium concentrations (so they sum to 1 over the two sides) of the reaction
$$ABCD \rightarrow A+ BCD+AB+ CD+ABC+D.$$

That makes sense.

 

[tstl]

What class is this for by the way? I'm asking because if we know the level of the class, we can perhaps help better by knowing what level of work is expected.

This is not class homework but an actual research problem. The molecule in question is a toy example of a polymer. Under the mildest of condition we assume it undergoes a single breakage at a bond but as the condition becomes harsher it can break simultaneously at multiple bonds but that is out of the scope of discussion here. We want to keep the model as simple as possible.

 

[tstl]

It seems most likely that OP is expected to compute the relative equilibrium concentrations (so they sum to 1 over the two sides) of the reaction
$$ABCD \rightarrow A+ BCD+AB+ CD+ABC+D.$$


It seems like an introductory level science course, so it seems doubtful that it would be that heavy on probability theory. They probably are just the relative concentrations.

Edited grammar.

So this is then the list of probabilities of the 7 species after one bond breaking:( un normalized). Is this OK docnet?

P(A) = P(BCD) = 0.1
P(AB) = P(CD) = 0.7
P(ABC) = P(D) = 0.3
P(ABCD) = (1 - 0.1) * (1-0.7) * (1-0.3) = 0.9 * 0.3 * 0.7 = 0.189

 

[docnet]

So this is then the list of probabilities of the 7 species after one bond breaking:( un normalized). Is this OK docnet?

P(A) = P(BCD) = 0.1
P(AB) = P(CD) = 0.7
P(ABC) = P(D) = 0.3
P(ABCD) = (1 - 0.1) * (1-0.7) * (1-0.3) = 0.9 * 0.3 * 0.7 = 0.189

I want to help you, but I suspect that my simple assumptions about the final concentrations are incorrect, and I believe we need to rethink what happens when we consider the possibility of cleaving at the different sites simultaneously. So far we haven't used combinatorial probability at all.

 

[docnet]

If we make a wild guess and assume this scenario follows the general rules of chemical equilibria:

The $K_1,K2,K_3$ are Equilibrium constants.

$ABCD \rightarrow A + BCD$.
$K_1 = \frac{[A][BCD]}{[ABCD]}\sim \frac{.1\cdot .1}{.9} = \frac{1}{90}$.

$ABCD \rightarrow AB + CD$.
$K_2 = \frac{[AB][CD]}{[ABCD]}\sim \frac{.7\cdot .7}{.3} = \frac{49}{30}$.

$ABCD \rightarrow ABC + D$.
$K_3 = \frac{[ABC][D]}{[ABCD]}\sim \frac{.3\cdot .3}{.7} = \frac{9}{70}$.

The combined reaction is
$3ABCD \rightarrow A + AB + ABC + BCD + CD + D$.
$K = K_1K_2K_3=\frac{7}{3000}=\frac{[A][AB][ABC][BCD][CD][D]}{[ABCD]^3}$.

The right hand side of the last line gives the relative concentrations between the reactant and the products. But I don't know if we can get any further with it.

 

[haruspex]

the dissociation probs are as follows say:
A:B = 0.1, B:C = 0.7, C:D = 0.3

The way I read that the probs are independent.
So we end up with
ABCD (1-.1)(1-.7)(1-.3)
A|BCD (.1)(1-.7)(1-.3)
AB|CD (1-.1)(.7)(1-.3)
ABC|D (1-.1)(1-.7)(.3)
A|B|CD (.1)(.7)(1-.3)
A|BC|D (.1)(1-.7)(.3)
A|B|C|D (.1)(.7)(.3)