Bernoulli, binomial and geometric mechanisms

Try

Repeated independent trials have success probability p=1/3. One question counts successes in four trials; another records the trial number of the first success. Both involve the same trial mechanism but different random variables and supports. Sketch the possible values before using a family name.

Predict

Predict whether the first-success trial can be zero. Under the convention here it cannot: G=1 means immediate success. A convention counting failures before success instead starts at zero and shifts the mean by one. The word geometric without a support convention is incomplete.

Experiment

For X counting successes in four trials, P(X=2)=C(4,2)(1/3)²(2/3)²=8/27. For G, P(G=3)=(2/3)²(1/3)=4/27 and P(G>3)=(2/3)³=8/27.

Variable	What is fixed?	Possible values
X	Four trials	0,1,2,3,4
G	Stop at first success	1,2,3,…

List sequences for the first probability and a single stopping pattern for the second. Their equal-looking powers serve different counting rules.

Understand

A Bernoulli variable is one binary trial, with probability p at one and 1−p at zero. A binomial variable sums n independent Bernoulli trials sharing p. Its mass is C(n,k)p^k(1−p)^(n−k). The coefficient counts success positions; the product gives the probability of one sequence under independence.

For the trial-index geometric variable, P(G=k)=(1−p)^(k−1)p for k≥1 and 0<p≤1. Survival P(G>k)=(1−p)^k is often simpler than the mass. The probabilities sum to one by the geometric series. At p=1, G is deterministically one; at p=0, a first success never occurs and the finite-valued law needs separate treatment.

Memorylessness follows from survival ratios: P(G>s+t|G>s)=(1−p)^(s+t)/(1−p)^s=(1−p)^t whenever the conditioning event has positive probability. The remaining trials retain the same independent success mechanism. This is a property of the model, not every real waiting process.

The tail-sum identity for positive integer G gives E(G)=Σ from k=0 to infinity P(G>k)=1/p. The failures-before-success variable G−1 therefore has mean (1−p)/p. Its support and mass formula shift too; mixing conventions is a common off-by-one error.

The number of trials until a fixed later success has a negative-binomial mechanism, while sampling a fixed number of objects without replacement from a finite population gives a hypergeometric count. These names help organize mechanisms, but the mechanisms should determine the law, not the presence of an integer outcome alone.

Changing trial probabilities or introducing dependence generally breaks the simple binomial and geometric formulas. A changing environment can make a constant-p assumption unrealistic even if each outcome is binary. Check that assumption before interpreting a long-run expectation or survival calculation.

For a binomial count, the mean np counts expected successes while variance np(1−p) quantifies repeated-count spread. A tail probability such as at least one success instead uses a union or complement; it is not obtained by reporting the mean.

Apply

For each practice problem, state the support and identify what is fixed: trials, successes or population composition. Explain the geometric expectation by a tail sum, not just a quoted formula. For a waiting-time story, identify a plausible reason memorylessness might fail.

Continue

Next consider counts with no fixed finite maximum. The Poisson distribution can arise as a rare-event limit, but one Poisson marginal does not establish a whole independent-increment process.

Continue with negative-binomial stopping and censoring for the full mass/normalization/moment argument, and multinomial counts and protocol dependence for joint allocations and covariance. These preserve the existing geometric and without-replacement practice.