19.3: Distribution of Results for Multiple Trials with Many Possible Outcomes
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It is now easy to extend our results to multiple trials with any number of outcomes. Let the outcomes be \(A\), \(B\), \(C\), …., \(Z\), for which the probabilities in a single trial are \(P_A\), \(P_B\), \(P_C\),…\(P_Z\). We again want to write an equation for the total probability after \(n\) trials. We let \(n_A\), \(n_B\), \(n_C\),…\(n_Z\) be the number of \(A\), \(B\), \(C\),…, \(Z\) outcomes exhibited in \(n_A+n_B+n_C+...+n_Z=n\) trials. If we do not care about the order in which the outcomes are obtained, the probability of \(n_A\), \(n_B\), \(n_C\),…, \(n_Z\) outcomes in \(n\) trials is
\[C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z \nonumber \]
and the total probability sum is
\[1={\left(P_A+P_B+P_C+\dots +P_Z\right)}^n=\sum_{n_I}{C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z} \nonumber \]
where the summation is to be carried out over all combinations of integer values for \(n_A\), \(n_B\), \(n_C\),…, \(n_Z\) consistent with \(n_A+n_B+n_C+...+n_Z=n\).
Let one of the terms for \(n_A\) \(A\)-outcomes, \(n_B\) \(B\)-outcomes, \(n_C\) \(C\)-outcomes, …, \(n_Z\)\({}_{\ }\)\(Z\)-outcomes, be
\[\left(P_{A,a}P_{A,b}\dots P_{A,f}\right)\left(P_{B,g}P_{B,h}\dots P_{B,m}\right)\times \left(P_{C,p}P_{C,q}\dots P_{C,t}\right)\dots \left(P_{Z,u}P_{Z,v}\dots P_{Z,z}\right) \nonumber \]
where there are \(n_A\) indices in the set \(\{a,\ b,\ \dots ,\ f\}\), \(n_B\) indices in the set \(\{g,\ h,\ \dots ,\ m\}\), \(n_C\) indices in the set \(\{p,\ q,\ \dots ,\ t\}\), …, and \(n_Z\) indices in the set \(\{u,\ v,\ \dots ,\ z\}\). There are \(n_A!\) ways to order the \(A\)-outcomes, \(n_B!\) ways to order the \(B\)-outcomes, \(n_C!\) ways to order the \(C\)-outcomes, …, and \(n_Z!\) ways to order the \(Z\)-outcomes. So, there are \(n_A!n_B!n_C!\dots n_Z!\) ways to order \(n_A\) \(A\)-outcomes, \(n_B\) \(B\)-outcomes, \(n_C\) \(C\)-outcomes, …, and \(n_Z\) \(Z\)-outcomes. The same is true for any other distinguishable combination; for every distinguishable combination belonging to the population set \(\{n_A\), \(n_B\), \(n_C\),…, \(n_Z\}\) there are \(n_A!n_B!n_C!\dots n_Z!\) indistinguishable permutations. Again, we can express this result as the general relationship:
total number of permutations = (number of distinguishable combinations)\({}_{\ }\)\({}_{\times }\) (number of indistinguishable permutations for each distinguishable combination)
so that
\[n!=n_A!n_B!n_C!\dots n_Z!C\left(n_A,n_B,n_C,\dots ,n_Z\right) \nonumber \]
and \[C\left(n_A,n_B,n_C,\dots ,n_Z\right)=\frac{n!}{n_A!n_B!n_C!\dots n_Z!} \nonumber \]
Equivalently, we can construct a sum, \(T\), in which we add up all of the \(n!\) permutations of \(P_{A,a}\) factors for \(n_A\) \(A\)-outcomes, \(P_{B,b}\) factors for \(n_B\) \(B\)-outcomes, \(P_{C,c}\) factors for \(n_C\) \(C\)-outcomes, …, and \(P_{Z,z}\) factors for \(n_Z\) \(Z\)-outcomes. The value of each term in \(T\) will be \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\). So we have
\[T=n!P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z \nonumber \]
\(T\) will contain all \(C\left(n_A,n_B,n_C,\dots ,n_Z\right)\) of the \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products (distinguishable combinations) that are a part of the total-probability sum. Moreover, \(T\) will also include all of the \(n_A!n_B!n_C!\dots n_Z!\) indistinguishable permutations of each of these \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products. Then we also have
\[T=n_A!n_B!n_C!\dots n_Z!C\left(n_A,n_B,n_C,\dots ,n_Z\right) \nonumber \] \[\times P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z \nonumber \]
Equating these two expressions for\(\ T\) gives us the number of \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\)-valued products
\[n!P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z=n_A!n_B!n_C!\dots n_Z! \nonumber \] \[\times C\left(n_A,n_B,n_C,\dots ,n_Z\right)P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z \nonumber \]
and hence,
\[C\left(n_A,n_B,n_C,\dots ,n_Z\right)=\frac{n!}{n_A!n_B!n_C!\dots n_Z!} \nonumber \]
In the special case that \(P_A=P_B=P_C=\dots =P_Z\), all of the products \(P^{n_A}_AP^{n_B}_BP^{n_C}_C\dots P^{n_Z}_Z\) have the same value. Then, the probability of any set of outcomes, \(\{n_A,n_B,n_C,\dots ,n_Z\}\), is proportional to \(C\left(n_A,n_B,n_C,\dots ,n_Z\right)\).