Try
A shared device is either available, state A, or busy, state B, at each daily inspection. We will use a synthetic assumed model, not measured transition data: an available device is available again next inspection with probability 0.80; a busy device becomes available with probability 0.30. A probability describes uncertainty, not a promise about the next individual device.
Suppose 40% are available now. Predict tomorrow’s available proportion. Simply averaging 0.80 and 0.30 would give 0.55, but it assigns half the starting devices to each state. The actual starting weights are 0.40 and 0.60. The correct weighted calculation is 0.40·0.80+0.60·0.30=0.32+0.18=0.50.
Write the transition matrix P with rows=current state and columns=next state:
| Current state | Next A | Next B | Row sum |
|---|
| A | 0.80 | 0.20 | 1 |
| B | 0.30 | 0.70 | 1 |
Each row is a conditional distribution. The first row describes what happens given that the current state is A; it does not describe the current proportion in A. Nonnegative entries and row sums of one are mathematical validity checks. They do not establish that a real device follows these probabilities.
Predict
Let p be the current probability of A, so 1−p is the probability of B. The next A probability is
p next=0.80p+0.30(1−p)=0.30+0.50p.
Our row distribution is π=(p,1−p), and the next distribution is πP. The same model can use a column distribution, but then its update matrix must be P transposed. Do not switch conventions halfway through a calculation.
Predict the next two rows before revealing them:
| Inspection step | A probability | B probability |
|---|
| 0 | 2/5 | 3/5 |
| 1 | 1/2 | 1/2 |
| 2 | 11/20 | 9/20 |
The second forecast reweights the transition rows using the updated distribution: 0.50·0.80+0.50·0.30=0.55. Reusing the initial 40/60 weights forever would repeatedly calculate the first step instead of evolving the process. With 100 devices, 55 available is an expected count at step two, not a guaranteed integer outcome.
Now consider a different question: if the next available proportion is 55%, what previous distribution could have produced it? Predict whether solving backward must always give a valid probability.
Experiment
Keep A stays A=80, B moves to A=30, Forecast start=40 and Separate next-state target=55. The fields use whole-number percentages: 80 means 0.80. Choose five forecast steps, predict the first two, then select Reveal forecast and backcast. Inspect both weighted contributions in each trace row. The sum of A and B remains exactly one.
The backcast target is independent of the starting distribution and forecast length. It asks for a predecessor one inspection before that target, not for an earlier point in the displayed forecast. With target 55, solve 0.55=0.30+0.50p to obtain p=0.50. Forward multiplication verifies the answer.
Change only the target to 90. Old results disappear until you reveal again. The algebra now gives p=(0.90−0.30)/0.50=1.20, and the other component is −0.20. Inspect the result: the matrix is invertible, but there is no valid probability distribution with those components. Clamping to 100% would produce next A=80%, not 90%; it changes the problem rather than repairing it.
Change both transition fields to 40. Try target 40, then target 50. Explain the different messages before reading the argument below. Reset restores the original controls and hides the result; temporary exploration does not award understanding or save a new learning record.
Understand
For general probabilities a=P(A next given A) and b=P(A next given B), the matrix is P=[[a,1−a],[b,1−b]]. Expanding the weighted update gives
p next=b+(a−b)p.
As p ranges from zero to one, this expression runs over the closed interval between a and b. Therefore a target outside [min(a,b),max(a,b)] is unreachable in one step. This geometric interval check is independent of solving the inverse equation.
The determinant is a(1−b)−b(1−a)=a−b. If a differs from b, the only algebraic predecessor is p=(target−b)/(a−b). It is a feasible probability exactly when 0≤p≤1. The previous B probability is 1−p, and forward substitution is a separate verification. The workbench keeps exact rational arithmetic for its hundredth-grid inputs, so singularity and feasibility do not depend on a guessed decimal tolerance.
If a=b, both rows are identical and every previous distribution gives the same next A probability b. For a=b=0.40, target 0.40 has infinitely many feasible predecessors. Target 0.50 has none. The model has discarded previous-distribution information; “inverse unavailable” must distinguish these two cases.
Backcasting can also amplify uncertainty. With a=0.51 and b=0.50, targets 0.50 and 0.51 correspond to previous probabilities zero and one. A one-percentage-point target change spans the whole prior range. Exact arithmetic does not make an uncertain observation exact. These calculations reconstruct aggregate distributions under an assumed law; they are not Bayesian probabilities of an individual device’s past state.
A stationary distribution reproduces itself after an update. For the original model, solve p=0.30+0.50p to obtain p=0.60. Subtract this value from both sides: p next−0.60=0.50(p−0.60). Repeated substitution gives p at step n=0.60+(p at step zero−0.60)(0.50)ⁿ. This proves convergence without calculus because powers of 0.50 tend to zero.
Stationarity alone does not prove convergence. The flip matrix with a=0,b=1 gives p next=1−p: a start of 0.40 alternates 0.40,0.60,0.40,0.60 rather than approaching its stationary value 0.50. The identity a=1,b=0 instead preserves every starting distribution. Even a stationary ensemble may contain individual devices that continue switching states.
Apply
Try a changed model with a=0.70,b=0.20 and start 0.60. Calculate two forecast steps and a predecessor for target 0.40 before entering the controls. Keep the one-step target separate from the two-step forecast.
Check the changed model
The recurrence is p next=0.20+0.50p. Forecasts from 0.60 are 0.50 and 0.45. The predecessor for target 0.40 is (0.40−0.20)/0.50=0.40. This prior is not the forecast’s starting 0.60. The one-step reachable interval is [0.20,0.70], so target 0.10 would be impossible; its algebraic predecessor is −0.20.
Write a short model critique. Markov means that after conditioning on the current state, earlier history does not further change the next-state probabilities. Time homogeneous means the same transition probabilities apply at each inspection. If busy devices that have already waited a long time behave differently from newly busy devices, the two-state Markov description may omit relevant history. If morning and evening transitions differ, a single fixed matrix may be inappropriate even with sufficient states.
The inspection interval matters: a daily transition matrix is not automatically an hourly one. Two daily updates give P² and a two-day forecast; they do not justify relabeling one update. Empirical transition counts would also have estimation uncertainty, and a state with no observed departures would not supply an estimated conditional row. The widget assumes its entries rather than claiming to infer them from evidence.
Continue
Retain the state definitions, time interval, matrix convention, exact weighted trace, proposed target, predecessor classification and forward check in your written response. Record which conclusions are mathematical consequences of the model and which assumptions still need observations.
For an accessible algebra refresher, revisit matrix laws, zero and identity. The existing stochastic state-model investigation extends the discussion to model critique and learner-designed states. This bridge is Unit 4 enrichment rather than examined AP Precalculus content; its work does not alter the exam-practice score.
