Conditional Binomial Laws From Multinomial Counts

Why Conditioning Changes the Count Model

In a multinomial experiment, each trial is assigned to exactly one of several categories. If we learn how many trials fell in one category, the number still available for the other categories is fixed. The count in a second category therefore has a binomial distribution among those remaining trials, but its success probability must be adjusted: it is the probability of that category given that a trial did not fall in the conditioned category.

This result is useful when a problem gives partial information about category counts and asks for a probability or expected count under that information. It is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let \((N_1,\ldots,N_k)\) be category counts from \(n\) independent trials, each of which falls in exactly one of \(k\) categories. Suppose each trial has category probabilities \(p_1,\ldots,p_k\), where \(p_l\geq 0\) and \(\sum_{l=1}^k p_l=1\). Then the count vector has a multinomial distribution with parameters \(n\) and \((p_1,\ldots,p_k)\). For distinct categories \(i\) and \(j\), we study the conditional distribution of \(N_j\) given \(N_i=r\).

The Conditional Binomial Law

Assume first that \(p_i<1\), that \(r\) is an integer with \(0\leq r\leq n\), and that \(P(N_i=r)>0\). Once \(N_i=r\) is known, exactly \(n-r\) trials are in categories other than \(i\). For any one of those trials, the conditional probability of category \(j\) is \(p_j/(1-p_i)\). The denominator is the total probability of not being in category \(i\).

Theorem: Under the conditions above, \(N_j\mid(N_i=r)\) has a binomial distribution with number of trials \(n-r\) and success probability \(q=p_j/(1-p_i)\). Thus, for an integer \(b\),

P(N_j=b\mid N_i=r) = \binom{n-r}{b} \left(\frac{p_j}{1-p_i}\right)^b \left(1-\frac{p_j}{1-p_i}\right)^{n-r-b}.

When \(q=0\), the conditional count is always zero. When \(q=1\), it is always \(n-r\). If \(0<q<1\), its support is the full range \(0,1,\ldots,n-r\). More generally, the possible values are among \(0,1,\ldots,n-r\); the support is a singleton at either boundary \(q=0\) or \(q=1\). The formula is not used with a zero denominator: if \(p_i=1\), the only possible conditioned count is \(r=n\), and every other category count is zero with conditional probability one.

Why the Formula Holds

A direct derivation uses the fact that categories \(i\), \(j\), and “all categories other than \(i\) and \(j\)” form a three-category multinomial model. Write \(m=n-r\), and let \(s\) be the count in the pooled remaining categories. Given \(N_i=r\), the count in category \(j\) is \(b\) precisely when \(s=m-b\). For \(p_i<1\), the marginal probability of \(N_i=r\) is binomial with parameters \(n,p_i\). Dividing the relevant three-category probability by this marginal probability gives

\begin{aligned} P(N_j=b\mid N_i=r) &= \frac{\dfrac{n!}{r!\,b!\,(m-b)!} p_i^r p_j^b(1-p_i-p_j)^{m-b}} {\dfrac{n!}{r!\,m!}p_i^r(1-p_i)^m}\\ &= \binom{m}{b} \left(\frac{p_j}{1-p_i}\right)^b \left(\frac{1-p_i-p_j}{1-p_i}\right)^{m-b}. \end{aligned}

Since \((1-p_i-p_j)/(1-p_i)=1-p_j/(1-p_i)\), this is the binomial probability mass function in the theorem. The calculation also explains the two changes caused by conditioning: the trial count becomes \(n-r\), and the probability for category \(j\) is renormalized among trials known not to be in category \(i\).

The standard binomial mean and variance therefore give

E[N_j\mid N_i=r]=(n-r)q,\qquad \operatorname{Var}(N_j\mid N_i=r)=(n-r)q(1-q), \qquad q=\frac{p_j}{1-p_i}.

These formulas include the boundary cases: a count that is constant has conditional variance zero. They apply only when the conditioning event has positive probability; in particular, \(r\) must be a possible count under the stated multinomial model.

Worked Example: A Conditional Count in a Three-Outcome Survey

In a hypothetical survey, 12 independent responses are classified as type A, type B, or other, with probabilities \(0.25\), \(0.30\), and \(0.45\), respectively. Given that exactly 4 responses are type A, find the conditional probability of exactly 3 type B responses and the conditional expected number of type B responses.

Method. Apply the conditional binomial law. The conditioned category is A, so 4 of the 12 trials are already assigned and 8 remain. Among the remaining trials, the conditional probability of type B is

q=\frac{0.30}{1-0.25}=\frac{0.30}{0.75}=0.40.

Thus \(N_B\mid(N_A=4)\) is binomial with \(8\) trials and success probability \(0.40\). For exactly 3 type B responses,

\begin{aligned} P(N_B=3\mid N_A=4) &=\binom{8}{3}(0.40)^3(0.60)^5\\ &=56(0.064)(0.07776)\\ &=0.27869184\\ &\approx 0.2787. \end{aligned}

The conditional expectation is \(8(0.40)=3.2\) type B responses. As a check, both calculations use the same conditional model: 8 remaining trials, not 12. The common mistake avoided here is to use \(0.30\) as the conditional probability; among responses known not to be A, the correct probability is \(0.30/0.75=0.40\).

Calculating a Conditional Range Probability

For an event involving a range of counts, add the conditional binomial probabilities for all counts in that range. Before calculating, confirm that every count in the sum is possible and that the number of trials is the number left after conditioning.

Worked Example: A Range of Counts After Conditioning

A hypothetical quality-control check classifies each of 10 components as having a coating flaw (A), a seal flaw (B), or neither, with respective probabilities \(0.20\), \(0.30\), and \(0.50\). Given that exactly 3 components have a coating flaw, find the probability that between 2 and 4 components, inclusive, have a seal flaw. Also find the conditional mean and variance of the seal-flaw count.

Method. There are \(10-3=7\) components not classified as A. For any such component, the conditional probability of B is

q=\frac{0.30}{1-0.20}=\frac{0.30}{0.80}=\frac{3}{8}.

Therefore \(N_B\mid(N_A=3)\) is binomial with parameters \(7\) and \(3/8\). The requested probability is the sum of the probabilities for counts 2, 3, and 4:

\begin{aligned} P(2\leq N_B\leq4\mid N_A=3) &=\sum_{b=2}^{4}\binom{7}{b} \left(\frac{3}{8}\right)^b \left(\frac{5}{8}\right)^{7-b}\\ &=\frac{590625+590625+354375}{2097152}\\ &=\frac{1535625}{2097152}\\ &\approx 0.7322. \end{aligned}

The conditional mean is \(7(3/8)=21/8=2.625\). The conditional variance is

7\left(\frac{3}{8}\right)\left(\frac{5}{8}\right) =\frac{105}{64}=1.640625.

The probabilities in the sum are approximately \(0.2816\), \(0.2816\), and \(0.1690\), which total \(0.7322\) to four decimal places. The conditional mean also lies within the calculated range, a reasonable check but not a proof of the probability calculation. The mistake avoided is counting all 10 components as trials for B after 3 are already known to be A.

Boundary Cases and Support

The conditional binomial description includes degenerate cases, but the support must reflect them. If \(p_j=0\), category \(j\) cannot occur, so the conditional count is zero. If \(p_j=1-p_i\), every trial outside category \(i\) must be in category \(j\), so the conditional count is \(n-r\). Only when \(0<p_j<1-p_i\) can every integer from 0 through \(n-r\) occur with positive conditional probability.

Worked Example: Recognizing a Singleton Conditional Support

In a hypothetical inspection of 8 items, each item is classified as either A or B, with probabilities \(0.20\) and \(0.80\); there is no third outcome. Given that exactly 3 items are A, find the conditional distribution of the number of B items, its expectation, and the probability of exactly 4 B items.

Method. Here \(p_A=0.20\), \(p_B=0.80=1-p_A\), and \(n-r=8-3=5\). The conditional success probability is

q=\frac{0.80}{1-0.20}=1.

A binomial count with 5 trials and success probability 1 is always 5. Thus the conditional distribution assigns probability 1 to \(N_B=5\), and its support is the singleton \(\{5\}\). Consequently,

E[N_B\mid N_A=3]=5(1)=5,\qquad P(N_B=4\mid N_A=3)=0.

The count check is \(3+5=8\), as required. The common mistake avoided is reporting the full range \(0,1,\ldots,5\) just because 5 trials remain; at the boundary \(q=1\), only the value 5 is possible.

Common Mistakes to Avoid

Using the original sample size. Given \(N_i=r\), the conditional binomial trial count is \(n-r\), not \(n\).
Using the unadjusted category probability. The conditional success probability is \(p_j/(1-p_i)\), not \(p_j\), because the conditioned trials are known not to be in category \(i\).
Ignoring possible values. The count must be an integer from 0 through \(n-r\), and at a boundary conditional probability it may have only one possible value.
Conditioning on an impossible event. A conditional distribution given \(N_i=r\) requires \(P(N_i=r)>0\). For example, if \(p_i=0\), only \(r=0\) is possible.
Forgetting that the remaining categories are pooled. The denominator \(1-p_i\) includes every category other than \(i\), not only category \(j\).

Key takeaway: Given \(N_i=r\), there are \(n-r\) trials left outside category \(i\). For \(p_i<1\) and a possible conditioning event, the count in category \(j\) is binomial with probability \(p_j/(1-p_i)\); check whether that probability is at a boundary before stating the support.

Check Your Understanding

For each problem, identify the conditional binomial parameters before calculating.

A batch of 8 independently selected objects is classified as type A, type B, or other, with probabilities \(0.25\), \(0.50\), and \(0.25\). Given that exactly 2 are type A, find the probability that exactly 2 are type B. (Answer: \(0.0823\))
In 12 independent trials, category A has probability \(0.40\) and category B has probability \(0.30\); the remaining probability is assigned to other outcomes. Given that \(N_A=5\), find the conditional mean and variance of \(N_B\). (Answer: mean \(3.5\), variance \(1.75\))
In 9 independent trials, categories A and B each have probability \(0.20\), with probability \(0.60\) for other outcomes. Given \(N_A=3\), find \(P(N_B=0\mid N_A=3)\). (Answer: \(0.1780\))
Each of 6 trials is classified as A with probability \(0.50\) or B with probability \(0.50\). Given \(N_A=6\), state the conditional support of \(N_B\) and its expected value. (Answer: support \(\{0\}\), expected value \(0\))
In 5 independent trials, the only categories are A and B, with probabilities \(0.30\) and \(0.70\). Given \(N_A=1\), find the conditional probability that \(N_B=4\). (Answer: \(1\))

What You'll Learn