Binomial Distribution: Properties, Formulas, and Exam P Applications

Definition and Parameters

The binomial distribution models the number of successes X in n independent Bernoulli trials, each with success probability p. The PMF is P(X = k) = C(n, k) * p^k * (1-p)^(n-k) for k = 0, 1, ..., n. The mean is E[X] = np and the variance is Var(X) = np(1-p).

In actuarial contexts, "success" might mean a policyholder files a claim. If 20 policies each have a 0.1 probability of claim independently, then the number of claims follows Binomial(20, 0.1) with mean 2 and variance 1.8.

The four conditions for a binomial distribution are: (1) a fixed number of trials n, (2) each trial has exactly two outcomes (success or failure), (3) the probability of success p is the same for each trial, and (4) the trials are independent. If any of these conditions fails, the binomial model is not appropriate. For example, sampling without replacement from a finite population violates condition (4), and the hypergeometric distribution should be used instead.

Key Properties

The moment generating function (MGF) of X ~ Binomial(n, p) is M(t) = (1 - p + p*e^t)^n. This is useful for proving that the sum of independent binomial random variables with the same p is also binomial. If X ~ Bin(m, p) and Y ~ Bin(n, p) independently, then X + Y ~ Bin(m+n, p).

The mode of a binomial is the most likely value, approximately floor((n+1)p). The distribution is symmetric when p = 0.5 and becomes increasingly skewed as p moves away from 0.5.

The factorial moments of the binomial are especially clean. E[X(X-1)] = n(n-1)p^2, E[X(X-1)(X-2)] = n(n-1)(n-2)p^3, and so on. From the second factorial moment, you can quickly derive E[X^2] = n(n-1)p^2 + np, and therefore Var(X) = n(n-1)p^2 + np - (np)^2 = np(1-p).

The binomial distribution has a reproductive property: if X1, X2, ..., Xk are independent with Xi ~ Bin(ni, p), then their sum follows Bin(n1 + n2 + ... + nk, p). Crucially, the success probability p must be the same for all components. If the probabilities differ, the sum is not binomial.

Computing Binomial Probabilities

For small n, compute probabilities directly using the PMF. A useful recursive relationship is P(X = k+1) / P(X = k) = (n-k) * p / ((k+1) * (1-p)). This ratio formula lets you build up the entire PMF efficiently once you compute P(X = 0) = (1-p)^n.

For cumulative probabilities, use P(X <= k) = sum from j = 0 to k of C(n,j) * p^j * (1-p)^(n-j). When computing P(X >= k), it is often faster to use the complement: P(X >= k) = 1 - P(X <= k-1). For small k or small n-k, choose whichever sum has fewer terms.

Example: An insurer has 5 independent policies, each with a 0.3 probability of claim. Find P(at least 2 claims). P(X >= 2) = 1 - P(X = 0) - P(X = 1) = 1 - (0.7)^5 - 5(0.3)(0.7)^4 = 1 - 0.16807 - 0.36015 = 0.47178.

Normal Approximation

For large n, the normal approximation X is approximately Normal(np, np(1-p)) can be used when np >= 5 and n(1-p) >= 5. With the continuity correction, P(X >= k) is approximately P(Z >= (k - 0.5 - np) / sqrt(np(1-p))). Practice both exact and approximate calculations, as Exam P tests both.

The continuity correction improves accuracy because we are approximating a discrete distribution with a continuous one. The integer value k occupies the interval (k - 0.5, k + 0.5) on the continuous scale. So P(X >= k) for the discrete binomial corresponds to P(Y >= k - 0.5) for the continuous normal Y. Similarly, P(X = k) corresponds to P(k - 0.5 < Y < k + 0.5).

Example: If X ~ Bin(100, 0.4), find P(X >= 45) using the normal approximation. Here np = 40 and np(1-p) = 24. With continuity correction, P(X >= 45) is approximately P(Z >= (44.5 - 40) / sqrt(24)) = P(Z >= 4.5/4.899) = P(Z >= 0.918) = 1 - 0.8212 = 0.1788.

Poisson Approximation

When n is large and p is small, the Binomial(n, p) distribution is well approximated by Poisson(np). This approximation is accurate when n >= 50 and p <= 0.05 (though some references use np <= 10 as the guideline). The Poisson approximation is often simpler because the Poisson PMF involves only lambda = np, avoiding the computation of large binomial coefficients.

Example: Among 1000 policies, each has a 0.002 probability of a catastrophic claim. The number of catastrophic claims is approximately Poisson(2). The probability of at least one catastrophic claim is 1 - e^(-2) = 0.8647.

Common Pitfalls

A common mistake is using the binomial when trials are not independent. If you draw cards without replacement, each draw changes the probabilities for subsequent draws, violating the binomial assumption. Use the hypergeometric distribution in those cases.

Another error is forgetting to check whether the success probability is constant across trials. If different policies have different claim probabilities, the total number of claims is not binomial. In that case, you would need to use a Poisson binomial distribution or approximate with a Poisson or normal distribution.

Study Tips

For Exam P, memorize the PMF, mean, variance, and MGF of the binomial distribution. Practice computing exact probabilities for n up to 10 and using the normal or Poisson approximation for larger n. Know the conditions for each approximation and when continuity corrections are appropriate. About three to five binomial problems appear on each Exam P administration, so fluency with this distribution is essential. Build speed by practicing the complement approach for "at least" problems and the recursive ratio formula for building PMF tables.