Probability | Poisson Distribution | At Least One Error

Question Suppose that the number of broken threads on a given T-shirt of a line of clothing is a Poisson distribution with parameter \(\lambda = 1\). Calculate the probability that there is at least 1 one broken thread.

• We will solve the problem numerically.
• Perform a simulation to get a realization of the mean reaching \(\lambda\) in the long run.
• Visualize results 😉

Reference: Sheldon Ross Probability Modelling Edition 10

Step 01: Numerical Solution

We need the probability that the number of errors is at-least one. \[P\{X \ge 1\} = 1 - P\{X = 0\}\] \[P\{X \ge 1\} = 1 - e^{-\lambda} \frac{\lambda^{0}}{0!}; \lambda = 1\] \[P\{X \ge 1\} = 1 - e^{-1} \frac{1^{0}}{0!}\] \[P\{X \ge 1\} = 1 - \frac{1}{e}\] \[P\{X \ge 1\} \approx 0.632\]

Step 02: Simulation and Visualization

In each of the squares, the thin horizontal lines are the number of events that happened in that second. The red line tracks mean as the events accumulate. See how the mean reaches the equilibrium gradually (even though they start out in a rugged manner).

    plot[] := Module[{poisson, sample = 500, movingProbability, scale = 0.001},
    poisson = RandomVariate[PoissonDistribution[1], sample];
    movingProbability = Transpose[{Reverse[Accumulate[poisson]/Range[sample]], scale*Range[sample]}];
    Graphics[{
        Table[{Opacity[0.1], Line[{{0, scale*r}, {poisson[[r]], scale r}}]}, {r, 1, sample}], {Red, Line[movingProbability]}},
        ImageSize -> 240,
        AspectRatio -> 1,
        Frame -> True,
        FrameTicks -> {Automatic, None},
        PlotRange -> {{0, 6}, All}]
    ]
    plot[number_] := Parallelize[Table[plot[], number]]
    Export[StringReplace[NotebookFileName[], ".nb" -> ".svg"], GraphicsGrid[Partition[plot[30], 3]], ImageSize -> 754, ImageResolution -> 600]