A set of $n$ people are given some sweets. There is candy A, B and C and each is given with probability $p_A,p_B$ and $p_C$.

I am trying to find the possible combinations of this system.

We can use the multinomial theorem, assuming that a fraction $i_X$ receives candy $X\in A,B,C$ such that:

$$ \sum_{i_{X}=n} \begin{pmatrix} n\\ i_A,i_B,i_{C}, i \end{pmatrix} p_A^{i_A} p_B^{i_B} p_C^{i_{C}} [1-p_A-p_B-p_{C})]^{i} $$ where the sum is over all the combinations of $i_X$ that sum to $n$. This leads to: $$ (p_A + p_B + p_C + (1-p_A-p_B-p_{C}))^n $$ However, I don't see why the following is also not a valid approach to this question.

$$ \sum^{n}_{i_A=0} \begin{pmatrix} n \\ i_A \end{pmatrix} p_A^{i_A} (1-p_A)^{n-i_{A}} \sum^{n-i_A}_{i_{B}=0} \begin{pmatrix} n-i_A \\ i_B \end{pmatrix} p_B^{i_{B}}(1-p_B)^{n-i_{A}-i_{B}} \sum^{n-i_{A}-i_{B}}_{i_{C}=0} \begin{pmatrix} n-i_{A}-i_{B}\\ i_{C} \end{pmatrix} p_{C}^{i_{C}}(1-p_{C})^{n-i_{A}-i_{B}-i_{C}} $$ which leads to $$ [p_A + (1-p_A)\big[ p_B + (1-p_B)[p_{C} +(1-p_{C})]\big]]^{n} $$ Which is essentially the binomial theorem used in succession for the success or fail of each candy.

What is the difference between each approach?

  • $\begingroup$ $(1-p_A)(1-p_B)(1-p_C)^i$ as factor is wrong. It should be $(1-p_A-p_B-p_C)^i$ (where $i_A+i_B+i_C+i=n$). Further the summation leads to $(p_A+p_B+p_C+(1-p_A-p_B-p_C))^n=1^n=1$ as it should. $\endgroup$ – drhab Jul 1 '18 at 11:49
  • $\begingroup$ If persons get more that one sweet then multionomial distribution cannot be practicized. $\endgroup$ – drhab Jul 1 '18 at 12:06
  • $\begingroup$ @drhab thank you for your comments, they are most helpful, I have modified the question! $\endgroup$ – RedPen Jul 2 '18 at 9:24

Let $X_A,X_B,X_C$ denote the number of persons that receive a candy of sort $A,B,C$ respectively. Then $$P(X_A=i_A,X_B=i_B,X_C=i_C)=\binom{n}{i_A,i_B,i_C,i}p_A^{i_A}p_B^{i_B}p_C^{i_C}(1-p_A-p_B-p_C)^i$$ where $i_A+i_B+i_C+i=n$.

We can also write $$P(X_A=i_A,X_B=i_B,X_C=i_C)=$$$$P(X_A=i_A)P(X_B=i_B\mid X_A=i_A)P(X_C=i_C\mid X_A=i_A, X_B=i_B)=$$$$\binom{n}{i_A}p_A^{i_A}(1-p_A)^{n-i_A}\times$$$$\binom{n-i_A}{i_B}\left(\frac{p_B}{1-p_A}\right)^{i_B}\left(1-\frac{p_B}{1-p_A}\right)^{n-i_A-i_B}\times$$$$\binom{n-i_A-i_B}{i_C}\left(\frac{p_C}{1-p_A-p_B}\right)^{i_C}\left(1-\frac{p_C}{1-p_A-p_B}\right)^{i}$$ where again $i_A+i_B+i_C+i=n$.

If under e.g. condition that $i_A$ persons receive candy of sort $A$ and we are focusing on the question how many under this condition will receive candy of sort $B$ then we are dealing with a binomial distribution with parameters $n-i_A$ and $\frac{p_B}{1-p_A}$ (not $p_B$ as you seem to suggest).


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