# Thread: Probability of Vigorous Reaction

1. ## Probability of Vigorous Reaction

A person carries 3 bottles of compound X, 3 bottles of compound Y and 3 bottles of compound Z. While X and Y will react vigorously, Z will prevent the reaction from happening. If the person falls and 3 random bottles react, what is the probability that a vigorous reaction will happen?

I tried solving this question like this:
Possible combinations:
ABB
ABA

3/6 x 3/5 x 2/4 = 18/120
3/6 x 3/5 x 2/4 = 18/120

18/120 x 2 = 3/10

I'm wondering if the order of the bottles matter and if the the number of bottles of Z should be counted (e.g. 3/9 x 3/8 x 2/7)?

Please tell me if I'm wrong and provide the right way to do this...

2. ## Re: Probability of Vigorous Reaction

Originally Posted by Estermont
A person carries 3 bottles of compound X, 3 bottles of compound Y and 3 bottles of compound Z. While X and Y will react vigorously, Z will prevent the reaction from happening. If the person falls and 3 random bottles react, what is the probability that a vigorous reaction will happen?
I tried solving this question like this:
Possible combinations:
ABB
ABA
Please tell me if I'm wrong and provide the right way to do this...
This is like solving $X+Y+Z=3$ in the non-negative integers.
That is $\dbinom{3+3-1}{3}=\dbinom{5}{3}=10$ solutions.
You rightly have noticed that just two of the solutions will cause a reaction.
Order does not play a role in any of this.
That least one each of $X~\&~Y$ must react but no $Z$.
So I make it two vigorous out of ten possible reactions.

3. ## Re: Probability of Vigorous Reaction

Originally Posted by Plato
This is like solving $X+Y+Z=3$ in the non-negative integers.
That is $\dbinom{3+3-1}{3}=\dbinom{5}{3}=10$ solutions.
You rightly have noticed that just two of the solutions will cause a reaction.
Order does not play a role in any of this.
That least one each of $X~\&~Y$ must react but no $Z$.
So I make it two vigorous out of ten possible reactions.
Can you explain how X + Y + Z = 3 comes into this? (a link with explanation might be helpful)

4. ## Re: Probability of Vigorous Reaction

Originally Posted by Plato
This is like solving $X+Y+Z=3$ in the non-negative integers.
That is $\dbinom{3+3-1}{3}=\dbinom{5}{3}=10$ solutions.
You rightly have noticed that just two of the solutions will cause a reaction.
Order does not play a role in any of this.
That least one each of $X~\&~Y$ must react but no $Z$.
So I make it two vigorous out of ten possible reactions.
The topic is known as multi-selections. To count the number of ways to put $k$ identical objects into $N$ different distinct cells is $\dbinom{N+k-1}{N}$.
How many solutions to $A+B+C+D+E+F=50$ using non-negative integers? Well we think of six different cells (disjoint containers), the vairables, into which we can put the fifty identical ones . Answer $\dbinom{50+6-1}{50}$.

An icecream shop has twenty-six flavors and a order consists of four scoops of any and all flavors with repetitions allowed. How many possible combinations can the advertise?
ANSWER: $\dbinom{4+26-1}{4}$. [The flavors are different & the selections are identical]