91影视

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

Consider the discrete random variables \({\rm{X and Y}}\)

\({\rm{p(x,y)}}\)is the joint probability mass function

\({\rm{p(x,y) = P(X = x,Y = y)}}\)

and \(\mathop {\rm{{aa}}}\limits_{\rm{x}} {\rm{n}}\mathop {\rm{{aa}}}\limits_{\rm{y}} {\rm{np(x,y) = 1}}\)for every pair \({\rm{(x,y)}}\)

(a) The following is true, as we can see from the definition above.

\({\rm{P(X = 1\;and\;Y = 1) = p(1,1) = 0}}{\rm{.2}}\)

where the value can be found in the supplied table

(b) The following vacations

\(\begin{aligned}{}{P(X拢1\;and\;Y拢1)\mathop = \limits^{(1)} p(0,0) + p(0,1) + p(1,0) + p(1,1)} \\{= 0.1 + 0.04 + 0.08 + 0.2} \\{ = 0.42.}\end{aligned}\)

(1): the occurrence

\(\{ X拢1\;and\;Y拢1\} \)

can be written as the sum of four separate events

\({\rm{\{ X = 0,Y = 0\} ,\{ X = 0,Y = 1\} ,\{ X = 1,Y = 0\} ,\{ X = 1,Y = 1\} }}{\rm{.}}\)

(c) Because neither random variable is equal to zero, at least one of the hoses is used on both islands.

The likelihood of the given event is as follows:

\(\begin{aligned}{P(X0\;and\;Y0)\mathop= \limits^{(2)} p(1,1) + p(1,2) + p(2,1) + p(2,2)} \\{ = 0.2 + 0.06 + 0.14 + 0.3}\\{ = 0.7.}\end{aligned}\)

(d) The mass function of marginal probability

random variable that is discrete \({\rm{X}}\) is

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\mathop {\rm{{aa}}}\limits_{\rm{y}} {\rm{np(x,y),}}\;\;\;{\rm{\;for every\;x}}\)

The mass function of marginal probability

random variable that is discrete \({\rm{Y}}\)is

\({{\rm{p}}_{\rm{Y}}}{\rm{(y) = }}\mathop {\rm{{aa}}}\limits_{\rm{x}} {\rm{np(x,y),}}\;\;\;{\rm{\;for every\;y}}{\rm{.}}\)

The following is true for \({\rm{x^I \{ 0,1,2\} }}\). according to the definition.

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\mathop {\rm{{aa}}}\limits_{{\rm{y\^I \{ 0,1,2\} }}} {\rm{np(x,y) = p(x,0) + p(x,1) + p(x,2)}}{\rm{.}}\)

The marginal probability is the sum of each row for every \({\rm{x^I \{ 0,1,2\} }}\). As a result, we've

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.1 + 0}}{\rm{.04 + 0}}{\rm{.02 = 0}}{\rm{.16,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.08 + 0}}{\rm{.2 + 0}}{\rm{.06 = 0}}{\rm{.34,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.06 + 0}}{\rm{.14 + 0}}{\rm{.3 = 0}}{\rm{.5,}}}\end{array}\)

which calculates \({\rm{X }}\)marginal \({\rm{pmf}}\)

Similarly, for \({\rm{Y}}\)marginal \({\rm{pmf}}\), we get

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{Y}}}{\rm{(0) = 0}}{\rm{.1 + 0}}{\rm{.08 + 0}}{\rm{.06 = 0}}{\rm{.24}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(1) = 0}}{\rm{.04 + 0}}{\rm{.2 + 0}}{\rm{.14 = 0}}{\rm{.38}}}\end{array}\)

\({{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.3 = 0}}{\rm{.38}}\)

where we add up the probabilities in each column

To summaries, we have \({\rm{X and Y}}\)marginal \({\rm{pmf}}\), respectively, supplied with

\(\begin{aligned}{}{{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.16}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.34}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.5,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(x) = 0,}}\;\;\;{\rm{x"I \{ 0,1,2\} }}{\rm{.}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(1) = 0}}{\rm{.38,}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.38,}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(y) = 0,}}\;\;\;{\rm{y"I \{ 0,1,2\} }}{\rm{.}}}\end{aligned}\)

Using \({\rm{X }}\)minimal \({\rm{pmf}}\), we get

\(P(X拢1) = {p_X}(0) + {p_X}(1) = 0.16 + 0.34 = 0.5.\)

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

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

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

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

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

Otherwise, they are reliant on the following factors:

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{X}}}{\rm{(2) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.5 \times 0}}{\rm{.38 = 0}}{\rm{.19}}}\\{{\rm{p(2,2) = 0}}{\rm{.3}}}\end{array}\)

It leads us to the conclusion that the following holds true for the pair \({\rm{(2,2)}}\)

\(\begin{array}{*{20}{c}}{p(2,2) = 0.30.19 = {p_X}(2) \times {p_Y}(2)} \\{p(2,2){p_X}(2) \times {p_Y}(2)}\end{array}\)

The random variables are so dependent.