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Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Random variable \({\rm{X}}\) with pdf

is said to have exponential distribution with parameter\({\rm{\lambda }}\).

Therefore, we have

and

(a):

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 \({\rm{Y}}\) 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.

We are given that random variables \({\rm{X}}\) and \({\rm{Y}}\) are independent, therefore, the joint pdf of \({\rm{X}}\) and \({\rm{Y}}\) is

(b):

As given in the hints, the following is true

\(\begin{aligned}{l}{\rm{P(X£1 and Y£1)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X£1) \times P(Y£1)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 1}}}}} \right){\rm{ \times }}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 1}}}}} \right)\\{\rm{ = 0}}{\rm{.4}}\end{aligned}\)

(1): from the multiplication property given below,

(2): cof of exponentially distributed random variable is

Multiplication Property: Two events A and B are independent if and only if

\({\rm{P(A\c{C}B) = P(A) \times P(B)}}\)

(c):

From the picture we can notice over what area should we integrate (the shaded area). Hence,

\(\begin{aligned}{\rm{P(X + Y£2)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \int_{\rm{0}}^{\rm{2}} {\int_{\rm{0}}^{{\rm{2 - x}}} {{{\rm{e}}^{{\rm{ - x - y}}}}} } {\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\int_{\rm{0}}^{{\rm{2 - x}}} {{{\rm{e}}^{{\rm{ - x}}}}} } {{\rm{e}}^{{\rm{ - y}}}}{\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{ - }}\left. {{{\rm{e}}^{{\rm{ - y}}}}} \right|_{\rm{0}}^{{\rm{2 - x}}}} \right){\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{x - 2}}}}} \right){\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {{{\rm{e}}^{{\rm{ - x}}}}} {\rm{dx - }}\int_{\rm{0}}^{\rm{2}} {{{\rm{e}}^{{\rm{ - 2}}}}} {\rm{dx}}\\{\rm{ = - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{0}}^{\rm{2}}{\rm{ - 2}}{{\rm{e}}^{{\rm{ - 2}}}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - 2}}}}{\rm{ - 2}}{{\rm{e}}^{{\rm{ - 2}}}}\\{\rm{ = 0}}{\rm{.594}}\end{aligned}\)

(1): for every adequate set \({\rm{A}}\) the following holds

d)

The following is true

\(\begin{aligned}{\rm{P(1£X + Y£2) = P(X + Y£2) - P(X + Y£1)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.594 - 0}}{\rm{.264}}\\{\rm{ = 0}}{\rm{.33}}\end{aligned}\)

(1): the first probability has been calculated. The second probability can be calculated the same way as the first, we get

\(\begin{aligned}{\rm{P(X + Y£1) = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - x}}} {{{\rm{e}}^{{\rm{ - x - y}}}}} } {\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - x}}} {{{\rm{e}}^{{\rm{ - x}}}}} } {{\rm{e}}^{{\rm{ - y}}}}{\rm{dydx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{ - }}\left. {{{\rm{e}}^{{\rm{ - y}}}}} \right|_{\rm{0}}^{{\rm{1 - x}}}} \right){\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{x - 1}}}}} \right){\rm{dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {{{\rm{e}}^{{\rm{ - x}}}}} {\rm{dx - }}\int_{\rm{0}}^{\rm{1}} {{{\rm{e}}^{{\rm{ - 1}}}}} {\rm{dx}}\\{\rm{ = - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{0}}^{\rm{1}}{\rm{ - }}{{\rm{e}}^{{\rm{ - 1}}}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - 1}}}}{\rm{ - }}{{\rm{e}}^{{\rm{ - 1}}}}\\{\rm{ = 0}}{\rm{.264}}{\rm{.}}\end{aligned}\)