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A discrete random variable is one that can only take on a finite number of different values.

A pmf with a random variable X is:

\({\rm{p(x;\mu ) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

Poisson Distribution with parameter \({\rm{\mu > 0}}\) is claimed to exist for \({\rm{x = 0,1,}}...\)

We're told that X has a Poisson Distribution with \({\rm{\mu = 0}}{\rm{.2}}\).

A=a disc is missing precisely one pulse.

The following statement is correct:

\(\begin{aligned}P(A) &= P(X = 1)\\ &= p(1;0,2)\\ &= {{\rm{e}}^{{\rm{ - 0}}{\rm{.2}}}}\frac{{{\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{1}}}}}{{{\rm{1!}}}}\\ &= {\rm{0}}{\rm{.1637}}\end{aligned}\)

you may determine this number using Appendix Table \({\rm{A}}{\rm{.2}}\), which contains the Poisson cdf \({\rm{F(x;\mu )}}\) as:

\(\begin{aligned}{\rm{F(1;0}}{\rm{.2) - F(0;0}}.2) &= 0{\rm{.982 - 0}}{\rm{.819}}\\ &= 0{\rm{.163}}\end{aligned}\)

Therefore, the value is: \({\rm{P(X = 1) = 0}}{\rm{.1637}}\).

A disc with at least two missing pulses is referred to as B.

The following statement is correct:

\(\begin{aligned}P(B) &= P(X \ge {\rm{2)}}\\ &= {\rm{1 - P(X}} \le {\rm{1)}}\\ &= 1 - F(1;0,2)\\ &= {\rm{1 - 0}}{\rm{.982}}\\ &= 0{\rm{.018}}\end{aligned}\)

(1): Because X can only take integer values,\({\rm{\{ X < 2\} }}\)is the same event as\({\rm{\{ X}} \le {\rm{1\} }}\);

(2):Table\({\rm{A}}{\rm{.2}}\)in Appendix.

Therefore, the value is: \({\rm{P(X}} \ge {\rm{2) = 0}}{\rm{.018}}\).

\({{\rm{C}}_{\rm{1}}}\)=the first disc has no missing pulses;

There is no missing pulse on the\({{\rm{C}}_{\rm{2}}}\)=second disc.

The possibilities of both outcomes are, of course, the same.

\(\begin{array}{c}{\rm{P(}}{{\rm{C}}_{\rm{1}}}{\rm{) = P(}}{{\rm{C}}_{\rm{2}}}{\rm{)}}\\{\rm{ = p(0;0}}{\rm{.2)}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - 0}}{\rm{.2}}}}\frac{{{\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{0}}}}}{{{\rm{0!}}}}\\{\rm{ = F(0;0,2)}}\\{\rm{ = 0}}{\rm{.819}}\end{array}\)

You have the option of using Appendix Table\({\rm{A}}{\rm{.2}}\)or calculating it yourself.

We are asked to calculate the chance of the two occurrences colliding.

\(\begin{array}{c}{\rm{P(}}{{\rm{C}}_{\rm{1}}} \cap {{\rm{C}}_{\rm{2}}}{\rm{)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(}}{{\rm{C}}_{\rm{1}}}{\rm{) \bullet P(}}{{\rm{C}}_{\rm{2}}}{\rm{)}}\\{\rm{ = 0}}{\rm{.819 \times 0}}{\rm{.819}}\\{\rm{ = 0}}{\rm{.671}}\end{array}\)