Probability | Poisson Distribution

Question

The number of coins that Henry spots when walking to park is a Poisson random variable with mean 6. Each coin is equally likely to be a penny, a nickel, a dime, or a quarter. Henry ignores the pennies but picks up the other coins.

Find the expected amount of money that Henry picks up on his way to park.
Find the variance of the amount of money that Henry picks up on his way to park.
Find the probability that Henry picks up exactly 25 cents on his way to park.

Solution | Analytical

There are three parts to the problem, and we can approach them systematically. Let us also assign symbols to the coin types for easier reference.

\(C_p\) is a penny
\(C_n\) is a nickel
\(C_d\) is a dime
\(C_q\) is a quarter

1. Expected amount of money

Expected amount of money is the easiest to solve amongst all the three questions. Let us tackle is systematically

All of the above coins \(\{C_p, C_n, C_d, C_q\}\) with the corresponding values of are equally likely to be found with a probability of \(\frac{1}{4}\)
Their values are \(C_p = 0¢, C_n = 5¢, C_d = 10¢, C_q = 25¢\)
The expected amount of money is \(E(N) E(C_x)\) where \(E(N)\) is the expected number of coins and E(C_x) is the expected amount per pick up.
\(E(n) = \lambda = 6\)
\(E(C_x) = \Sigma C_x p_x= \frac{1}{4} (0 + 5 + 10 + 25) = 10¢\)
The expected money is hence \(6 * 10¢ = 60¢\)

2. Variance of amount of money

\(Var(C_x) = E(N) * Var(C_x) + (E[C_x])^2 * Var(N)\) where \(N\) is the Poisson random variate with \(\lambda = 6\)

\(E(N) = 6\)
\(Var(X) = E[C_x^2] − (E[C_x])^2\)

\(E[C_x^2] = \frac{1}{4} (0^2 + 5^2 + 10^2 + 25^2) = \frac{750}{4} = 187.5\)
\((E[C_x])^2 = 10^2 = 100\)

\(E[C_x])^2 = 10 ^ 2 = 100\)
\(Var(N) = 6\)

Bringing all of the above together, \[Var(C_x) = E(N) * Var(C_x) + (E[C_x])^2 * Var(N) = 6 * 87.5 + 100 * 6 = 1125 \space ¢^2 \]

3. \(P(\Sigma x) = 25¢\)

We will solve this using simulation. Please scroll below

Solution | Simulation

1. Expected amount of money: The following code was used to calculate the mean money collected for \(x\) iterations

\(54.500 \rightarrow 10^1\) iterations
\(59.500 \rightarrow 10^2\) iterations
\(60.710 \rightarrow 10^3\) iterations
\(60.035 \rightarrow 10^4\) iterations
\(60.090 \rightarrow 10^5\) iterations
\(59.970 \rightarrow 10^6\) iterations
\(60.001 \rightarrow 10^7\) iterations

coinsCollected := RandomChoice[{0, 5, 10, 25}, RandomVariate[PoissonDistribution[6]]]

simulator[n_] := Module[{simulation}, simulation = Table[coinsCollected, n]; simulation]

Mean[Total /@ simulator[1000]] // N

2. Variance of amount of money: The following code was used to calculate the mean money collected for \(x\) iterations

\(0917.778 \rightarrow 10^1\) iterations
\(0961.806 \rightarrow 10^2\) iterations
\(1107.360 \rightarrow 10^3\) iterations
\(1123.720 \rightarrow 10^4\) iterations
\(1127.500 \rightarrow 10^5\) iterations
\(1122.080 \rightarrow 10^6\) iterations
\(1124.250 \rightarrow 10^7\) iterations

coinsCollected := RandomChoice[{0, 5, 10, 25}, RandomVariate[PoissonDistribution[6]]]

simulator[n_] := Module[{simulation}, simulation = Table[coinsCollected, n]; simulation]

Variance[Total /@ simulator[1000]] // N

3. \(P(\Sigma x) = 25¢\): The following code was used to probability for \(x\) iterations

\(0.000000 \rightarrow 10^1\) iterations
\(0.050000 \rightarrow 10^2\) iterations
\(0.047000 \rightarrow 10^3\) iterations
\(0.047100 \rightarrow 10^4\) iterations
\(0.045640 \rightarrow 10^5\) iterations
\(0.045284 \rightarrow 10^6\) iterations
\(0.045526 \rightarrow 10^7\) iterations

coinsCollected := RandomChoice[{0, 5, 10, 25}, RandomVariate[PoissonDistribution[6]]]

iterations = 10000000;

Length[Select[Total /@ simulator[iterations], # == 25 &]]/iterations // N

Visualization

        
ClearAll[coinsCollected, mapperText, mapperBar, mapperTotal, simulator]

totalTextPosX = 30;
rectOffset = 31;
xScaler = 0.05;

coinsCollected :=
 RandomChoice[{0, 5, 10, 25}, RandomVariate[PoissonDistribution[6]]]

mapperText[y_, list_List] := Module[{counts, keys, values},
  counts = Counts[list];
  MapThread[{{Red},
     Text[Style[#1, {Black, FontSize -> 10}], {#2, y}]} &, {Values@
     counts, Keys@counts}]
  ]
mapperBar[total_, y_] := {Opacity[0.5], Gray,
  Rectangle[{rectOffset, y - 0.05}, {rectOffset + xScaler*total,
    y + 0.1}]}
mapperTotal[total_,
  y_] := {Text[
   Style[StringPadRight[ToString[0.01*total], 4, "0"], {Black,
     FontSize -> 10}], {totalTextPosX, y}]}

simulator[n_] := Module[{simulation},
  simulation = Table[coinsCollected, n];
  simulation
  ]

plot := Module[{simulation, range, length, totals, a, b},
  simulation = simulator[100];
  totals = Total /@ simulation;
  length = Length@simulation;
  Echo[Mean@totals // N];
  range = Range[100/length, 100, 100/length];
  a = Graphics[{MapThread[mapperText[#1, #2] &, {range, simulation}],
     {White,
      Rectangle[{rectOffset, 0}, {rectOffset + xScaler*Max@totals,
        Max@range}]},
     MapThread[mapperTotal[#1, #2] &, {totals, range}],
     MapThread[mapperBar[#1, #2] &, {totals, range}]
     },
    PlotRange -> All, Frame -> True,
    FrameTicks -> {{{0, "x 1¢"}, {5, "x 5¢"}, {10, "x 10¢"}, {25,
        "x 25¢"}, {totalTextPosX, "Total$"}}, Automatic},
    GridLines -> {{0, 5, 10, 25}, Range[100]},
    GridLinesStyle -> {{Thickness[.02], Opacity[0.0],
       Green}, {Thickness[0], Gray, Opacity[0.1]}},
    FrameLabel -> {None, "Iterations"},
    ImageSize -> 800, AspectRatio -> 1.5]
  ]

Export[StringReplace[NotebookFileName[], {".nb" -> ".svg"}], plot]

Probability | Poisson Distribution | Many More Monies

Question

Solution | Analytical

Solution | Simulation

Visualization