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Thread: Probability of Vigorous Reaction

  1. #1
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    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...
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    Re: Probability of Vigorous Reaction

    Quote Originally Posted by Estermont View Post
    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.
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    Re: Probability of Vigorous Reaction

    Quote Originally Posted by Plato View Post
    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)
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    Re: Probability of Vigorous Reaction

    Quote Originally Posted by Plato View Post
    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]
    Thanks from Estermont
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