91影视

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

(a):

If and only if, two random variables \({\rm{X}}\)and \({\rm{Y}}\)are independent.

1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\)and when \({\rm{X}}\)and \({\rm{Y}}\)discrete rv's,

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\)and when \({\rm{X}}\)and \({\rm{Y}}\)continuous rv's,

otherwise they are dependent.

Given the independence of the random variables \({\rm{X}}\)and \({\rm{Y}}\), the joint pmf of \({\rm{X}}\)and \({\rm{Y}}\)is a product of marginal probability mass functions.

Reminder: A random variable \({\rm{X}}\)with pmf

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

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

Therefore, the joint pmf is

\(\begin{aligned}{\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{x}}}}{{{\rm{x!}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{y}}}}{{{\rm{y!}}}}\end{aligned}\)

\({\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{x}}{\rm{ \times \mu }}_{\rm{2}}^{\rm{y}}}}{{{\rm{x!y!}}}}\), \({\rm{x,y}} \in {\mathbb{N}_0}\)

\({\rm{p(x,y) = 0}}\), otherwise..

We can write it as

\({\rm{p(x,y)}} = \left\{ {\begin{aligned}{*{20}{l}}{{e^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{x}}{\rm{ \times \mu }}_{\rm{2}}^{\rm{y}}}}{{{\rm{x!y!}}}}}&{,x,y \in {\mathbb{N}_0},}\\0&{,{\rm{otherwise}}.}\end{aligned}} \right.\)

(b):

We should look at the sum of the two random variables to see if one or less error will be made on both tests combined. As a result, the following applies:

\(\begin{array}{l}{\rm{P(X + Y}} \le 1){\rm{ = p(0,0) + p(1,0) + p(0,1)}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{0}}{\rm{ \times \mu }}_{\rm{2}}^{\rm{0}}}}{{{\rm{0!0!}}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{1}}{\rm{ \times \mu }}_{\rm{2}}^{\rm{0}}}}{{{\rm{1!0!}}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{0}}{\rm{ \times \mu }}_{\rm{2}}^{\rm{1}}}}{{{\rm{0!1!}}}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\left( {{\rm{1 + }}{{\rm{\mu }}_{\rm{1}}}{\rm{ + }}{{\rm{\mu }}_{\rm{2}}}} \right).\end{array}\)

Therefore ,The probability is \({{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\left( {{\rm{1 + }}{{\rm{\mu }}_{\rm{1}}}{\rm{ + }}{{\rm{\mu }}_{\rm{2}}}} \right){\rm{;}}\)

(c):

The hint encapsulates all we'll be doing. We look at the total once more, so if \({\rm{m}}\)is a non-negative integer, we have

\(\begin{array}{l}{\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\rm{P}} {\rm{(X = k,Y = m - k)}}\\{\rm{ = }}\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}} \frac{{{\rm{\mu }}_{\rm{1}}^{\rm{k}}{\rm{ \times \mu }}_{\rm{2}}^{{\rm{m - k}}}}}{{{\rm{k!(m - k)!}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}{\rm{ \times }}\frac{{{\rm{m!}}}}{{{\rm{m!}}}}\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {{\rm{\mu }}_{\rm{1}}^{\rm{k}}} {\rm{ \times \mu }}_{\rm{2}}^{{\rm{m - k}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{k!(m - k)!}}}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{m!}}}}\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {{\rm{\mu }}_{\rm{1}}^{\rm{k}}} {\rm{ \times \mu }}_{\rm{2}}^{{\rm{m - k}}}{\rm{ \times }}\frac{{{\rm{m!}}}}{{{\rm{k!(m - k)!}}}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{m!}}}}\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {{\rm{\mu }}_{\rm{1}}^{\rm{k}}} {\rm{ \times \mu }}_{\rm{2}}^{{\rm{m - k}}}{\rm{ \times }}\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{m!}}}}{\rm{ \times }}{\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{ + }}{{\rm{\mu }}_{\rm{2}}}} \right)^{\rm{m}}}{\rm{,}}\end{array}\)

(1) : look at the given set \({\rm{A}}\). We can write event \({\rm{X + Y = m}}\)as union of \({\rm{m + 1}}\)disjoint events, therefore we get the sum.

(2) : we multiply with

\({\rm{1 = }}\frac{{{\rm{m!}}}}{{{\rm{m!}}}}\)

in order to obtain

\(\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right)\)

(3) : the Binomial Theorem (see the exercise).

The expression we received represents the pmf of a Poisson random variable with parameter \({{\rm{\mu }}_{\rm{1}}}{\rm{ + }}{{\rm{\mu }}_{\rm{2}}}\), as we can see from the definition of Poisson distribution. The total number of errors in the two exams follows the Poisson distribution described before.

Therefore,The probability is \({{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{m!}}}}{\rm{ \times }}{\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{ + }}{{\rm{\mu }}_{\rm{2}}}} \right)^{\rm{m}}}\).