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The expected value is calculated in statistics and probability analysis by multiplying each conceivable event by the probability that it will occur, and then summing all of those values. Investors can select the scenario that is most likely to result in the desired result by calculating anticipated values.

Proposition: The number of events during a time interval of length t can be modeled using Poisson Random Variable with parameter\(\mu = \alpha t\). This indicates that

\({P_k}(t) = {e^{ - \alpha t}} \cdot \frac{{{{(\alpha t)}^k}}}{{k!}}\)

It is also known as Poisson process (not formally defined).

Alpha given in the exercise

\(\alpha = 10\)

is the rate. We are also given the probability that the arriving vehicle has no equipment violations

\(p = 0.5\)

(a):

Define event

\(N = \{ {\rm{ no violations }}\} \)

Parameter\(\mu \)is, for\(t = 1\)hour

\(\mu = \alpha t = 10 \cdot 1 = 10\)

The following is true

\(\begin{array}{l}P(\{ X = 10\} \cap N)\mathop = \limits^{(1)} P(N\mid X = 10) \cdot P(X = 10)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop = \limits^{(2)} {0.5^{10}} \cdot (F(10;10) - F(9;10))\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop = \limits^{(3)} 0.00098 \cdot (0.583 - 0.458)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.00098 \cdot 0.125\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.00012\end{array}\)

(1): the multiplication rule given below;

(2): we have that

\(P(N\mid X = 10) = 0.5 \cdot 0.5 \cdot \ldots \cdot 0.5\)

where we are given the probability that a car, out of\(10\)cars, is\(0.5\); we also use the cdf of Poisson distribution-we also could have used the pmf;

(3): appendix Table A.2 contains the Poisson cdf\(F(x;10)\).

The Multiplication Rule

\(P(A \cap B) = P(A\mid B) \cdot P(B)\)

(b):

Similarly, as in (a), parameter\(\mu \)is, for\(t = 1\)hour

\(\mu = \alpha t = 10 \cdot 1 = 10\)

There are exactly\(10\)out of\(y \ge 10\)cars with no violations! We will use this to calculate the conditional probability below.

First, the following holds

\(\begin{array}{l}P(\{ y\;arrive\} \cap \{ 10\;with\;no\;violations\} )\\\mathop = \limits^{(1)} P(\{ 10\;with\;no\;violations\} \mid \{ y\;arrive\} \cdot P(\{ y\;arrive\} )\\\mathop = \limits^{(2)} P(10\;out\;of\;y\;vehicles\;have\;no\;violation) \cdot {e^{ - 10}} \cdot \frac{{{{10}^y}}}{{y!}}\end{array}\)

\(\begin{array}{l}\mathop = \limits^{(3)} \left( {\begin{array}{*{20}{c}}y\\{10}\end{array}} \right){0.5^{10}} \cdot {(1 - 0.5)^{y - 10}} \cdot {e^{ - 10}} \cdot \frac{{{{10}^y}}}{{y!}}\\ = {e^{ - 10}} \cdot \frac{{{5^y}}}{{10!(y - 10)!}};\end{array}\)

(1): here we use the multiplication rule given above;

(2): we have that X has Poisson distribution with parameter\(\mu = 10\), so

\(P(\{ y{\rm{ arrive }}\} ) = P(X = 10) = {e^{ - 10}} \cdot \frac{{{{10}^y}}}{{y!}}\)

(3): the probability that 10 out of y cars have no violation is

\(\left( {\begin{array}{*{20}{c}}y\\{10}\end{array}} \right){0.5^{10}} \cdot {(1 - 0.5)^{y - 10}}.\)

(c):

The following is true

\(\begin{array}{l}P(\{ 10{\rm{ cars withh no - violation }}\} )\mathop = \limits^{(1)} \sum\limits_{y = 10}^\infty {{e^{ - 10}}} \cdot \frac{{{5^y}}}{{10!(y - 10)!}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {e^{ - 10}} \cdot \frac{{{5^{10}}}}{{10!}}\sum\limits_{y = 10}^\infty {\frac{{{5^{y - 10}}}}{{(y - 10)!}}} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {e^{ - 10}} \cdot \frac{{{5^{10}}}}{{10!}}\sum\limits_{x = 0}^\infty {\frac{{{5^x}}}{{x!}}} \mathop = \limits^{(2)} {e^{ - 10}} \cdot \frac{{{5^{10}}}}{{10!}} \cdot {e^5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {e^{ - 5}} \cdot \frac{{{5^{10}}}}{{10!}}\end{array}\)

(1): event\(\{ 10\;cars\;with\;no - violation\} \)can be written as a union of all events

\(\{ y\;arrive\} \cap \{ 10with\,no\,violations\} \)

Where,\(y \ge 10\)because if there are less than\(10\)cars, it is impossible for exactly\(10\)to have no violation. Since the events are disjoint, we have sum of such events, where we have proved in (b) how does their probability actually look.

(2): it is well-known that

\(\sum\limits_{k = 0}^\infty {\frac{{{u^k}}}{{k!}}} = {e^u}\)

which we use for\(k = x\)and\(u = 5\).

Notice that

\({e^{ - 5}} \cdot \frac{{{5^{10}}}}{{10!}} = p(10;5)\)

This indicates that, when the number of cars that will not have violations (as in our case 10) within an hour has Poisson distribution with parameter \(5\). The number five we got from the given rate \(\alpha = 10\)and the probability \(p = 0.5\), therefore,

\(\alpha .p = 10 \cdot 0.5 = 5.\)