A Complete Introduction to Probability Investigation

Example The hub model has three submodels.

• Arrivals form a homogeneous Poisson process with rate $\lambda = 4$ per hour. The number $N(t)$ of arrivals in the first $t$ hours, and more generally the number in any time interval of length $t$ hours, has the $\operatorname{Pois}(4t)$ distribution, and the counts in disjoint intervals are independent.
• Each parcel is priority, the event $H$, with probability $0.25$. It has an upstream delay, the event $D$, with probability $P(D \mid H) = 0.3$ if it is priority and $P(D \mid H^{c}) = 0.1$ if not. The delay refers to travel before the hub, not to anything that happens at the hub. The priority and delay marks are independent across parcels and independent of the arrival times.
• A scanner is observed at hourly checkpoints $n = 0, 1, 2, \dots$, with checkpoint $n$ at hour $n$ on the arrival clock, and is either available, state $A$, or unavailable, state $U$. Its state $X_n$ is a Markov chain that starts at $X_0 = A$, and the chain is independent of the arrivals and the parcel marks taken together.

The transition matrix of the scanner chain, with rows and columns in the order $A$, $U$, is

Priority and ordinary parcels partition the parcels, and the two routes to a delay contribute $0.25 \cdot 0.3 = 0.075$ and $0.75 \cdot 0.1 = 0.075$. The law of total probability gives $P(D) = 0.15$, and Bayes’ theorem gives

A parcel is priority with probability $0.25$, but a delayed parcel is priority with probability $0.5$.

Example Let $W$ be the event that at least one parcel arrives between hour 2 and hour 2.25, the quarter hour after checkpoint 2. The arrival count for a quarter hour is $\operatorname{Pois}(1)$, since $4 \cdot 0.25 = 1$, so $P(W) = 1 - e^{-1}$. The card on checkpoint availability finds $P(X_2 = A) = 0.72$. The scanner chain and the arrivals were assumed independent, so

This is a statement about availability at one instant and arrivals in a later interval. It does not give the probability that every parcel arriving in that quarter hour is scanned at once, which would need the scanner to be available throughout the interval, something the model does not describe. By contrast, the product $P(H)\,P(D) = 0.25 \cdot 0.15 = 0.0375$ is not $P(H \cap D)$, which equals $0.075$, because priority and delay are dependent within one parcel.

Example For the next five parcels, $K \sim \operatorname{Bin}(5, 0.15)$, so $E[K] = 0.75$ and

For time, the chance of no arrivals in the first hour is $P(N(1) = 0) = e^{-4} \approx 0.0183$. From time $0$, the wait $T$ until the first arrival exceeds $t$ hours exactly when the first $t$ hours contain no arrival, so for $t \ge 0$

and $T \sim \operatorname{Exp}(4)$ with rate $4$ per hour. The same argument applied to the window after any fixed starting time gives the same distribution for the wait from that time. Fifteen minutes is $0.25$ hour, so the next arrival comes within 15 minutes with probability $1 - e^{-1} \approx 0.6321$. The mean wait is $1/4$ hour, which is 15 minutes, but the median wait is $(\ln 2)/4$ hour, about 10.4 minutes, and no single wait is guaranteed to be close to the mean. Delays in a one-hour window are a third question. Their expected number is $4 \cdot 0.15 = 0.6$ by the rule for random sums with independent counts, since the marks are independent of the count, but their distribution is not $\operatorname{Bin}(4, 0.15)$, because the number of parcels in the hour is random.

Example Starting from $X_0 = A$, the first checkpoint has $p_1 = 0.8$. A scanner available at checkpoint 2 got there either by staying available or by failing and recovering, so

Availability at both checkpoints 1 and 2 is a different event, with probability $0.8 \cdot 0.8 = 0.64$, because it excludes the path that fails and recovers. The deviation formula gives $p_2 = 2/3 + 0.16 \cdot 1/3 = 0.72$ again, and the later values $p_3 = 0.688$ and $p_4 = 0.6752$ continue toward $2/3$.

Example Raise the arrival rate from $4$ to $6$ per hour and keep everything else. The chance of no arrivals in an hour falls from $e^{-4} \approx 0.0183$ to $e^{-6} \approx 0.0025$, the mean wait falls from 15 to 10 minutes, and the chance of an arrival within 15 minutes rises from about $0.6321$ to $1 - e^{-1.5} \approx 0.7769$. The joint probability from the card on independence of submodels becomes

The five-parcel batch probability stays $1 - 0.85^{5} \approx 0.5563$, because a batch is defined by its parcels and not by time, and the scanner values $0.72$ and $2/3$ stay as well.

Example One run was recorded with Python 3.12.14 and the generator random.Random(55012). It simulated 10,000 independent five-parcel batches. For each parcel it drew the priority flag first, priority when random() was below $0.25$, and then the delay flag, delayed when the next draw was below $0.3$ for a priority parcel or below $0.1$ otherwise. At least one delay occurred in 5501 batches, an estimate of $0.5501$. The exact probability is about $0.5563$, and the standard error is

so the estimate lies about $1.25$ standard errors below the prediction, a gap of ordinary size for Monte Carlo error. The run also recorded 7443 delayed parcels among 50,000, a fraction of $0.1489$ against $P(D) = 0.15$, and 3745 of the delayed parcels were priority, a fraction of about $0.503$ against $P(H \mid D) = 0.5$. The run contains no arrival times and no scanner states, so it checks the parcel submodel and nothing else.