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A discrete variable function whose integral over any interval is the likelihood that the random variable it specifies will fall within that interval.

Let \({\rm{X}}\)and \({\rm{Y}}\)be discrete random variables.

The probability mass function combined

\({\rm{p(x,y)}}\)is

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

for every pair \({\rm{(x,y)}}\), where and \(\sum\limits_{\rm{x}} {\sum\limits_{\rm{y}} {\rm{p}} } {\rm{(x,y) = 1}}\).

The combined probability table is presented to us.

(a):

From the table, the following holds

\({\rm{P(X = 1,Y = 1) = p(1,1) = 0}}{\rm{.3}}\)(b):

The following is true

\(\begin{array}{l}{\rm{P(X}} \le {\rm{1 and Y}} \le 1){\rm{ = p(0,0) + p(0,1) + p(1,0) + p(1,1)}}\\{\rm{ = 0}}{\rm{.025 + 0}}{\rm{.015 + 0}}{\rm{.05 + 0}}{\rm{.03}}\\{\rm{ = 0}}{\rm{.12}}\end{array}\)

Therefore ,The probability is \({\rm{p(1,1) = 0}}{\rm{.3;}}\)

(b):

The following statement is correct:

Z Z \(\begin{array}{l}{\rm{P(X}} \le {\rm{1 and Y}} \le 1){\rm{ = p(0,0) + p(0,1) + p(1,0) + p(1,1)}}\\{\rm{ = 0}}{\rm{.025 + 0}}{\rm{.015 + 0}}{\rm{.05 + 0}}{\rm{.03}}\\{\rm{ = 0}}{\rm{.12}}\end{array}\)

Therefore ,The probability is \({\rm{P(X}} \le 1\)and \({\rm{Y}} \le {\rm{1) = 0}}{\rm{.12}}\).

(c):

The first requested probability is

\(\begin{array}{l}{\rm{P(X = 1) = p(1,0) + p(1,1) + p(1,2)}}\\{\rm{ = 0}}{\rm{.05 + 0}}{\rm{.03 + 0}}{\rm{.02}}\\{\rm{ = 0}}{\rm{.1}}\end{array}\)

The second probability requested is

\(\begin{array}{l}{\rm{P(Y = 1) = p(0,1) + p(1,1) + p(2,1) + p(3,1) + p(4,1) + p(5,1)}}\\{\rm{ = 0}}{\rm{.015 + 0}}{\rm{.03 + 0}}{\rm{.075 + 0}}{\rm{.09 + 0}}{\rm{.06 + 0}}{\rm{.03}}\\{\rm{ = 0}}{\rm{.3}}\end{array}\)

Therefore ,The probability is \({\rm{P(X = 1) = 0}}{\rm{.1}}\)

(d):

The likelihood of an overflow can be determined using the probability of event formula.

\({\rm{\{ X + 3Y > 5\} }}\)

as explained in the exercise.

Therefore ,We only take event for which \({\rm{(x,y),x + 3y}} \le {\rm{5}}\)is true.

(e):

Two random variables \({\rm{X}}\)and \({\rm{Y}}\)are independent if and only if

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 \(Y\)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.

We need to find marginal distributions to see if the random variables are independent.

The discrete random variable's marginal probability mass function is \({\rm{X}}\).

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\sum\limits_{\rm{y}} {\rm{p}} {\rm{(x:y): for every x,}}\)

The discrete random variable's marginal probability mass function is \({\rm{Y}}\).

\({{\rm{p}}_{\rm{Y}}}{\rm{(y) = }}\sum\limits_{\rm{x}} {\rm{p}} {\rm{(x,y), for every y}}{\rm{.}}\)

As we know, marginal probabilities for random variable \({\rm{X}}\) are calculated by summing probabilities in a certain row, while marginal probabilities for random variable \({\rm{Y}}\) are calculated by summing probabilities in a specific column.

The following is marginal pmf of \({\rm{X}}\)

And the marginal pmf of \({\rm{Y}}\)

Now we have to verify is for every \({\rm{(x,y)}}\), we have that

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

The following is the table of \({{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\)for every pair \({\rm{(x,y)}}\).

which is identical to the joint probability table.

As a result, the random variables are unrelated.