91Ó°ÊÓ

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.

(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. \({\rm{f(x,y) = }}\int_{\rm{X}} {{\rm{(x)}}} {\rm{ \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\), for covary \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) continuous rv's, otherwise they are dependent.

Since random variables \({\rm{X}}\) and \({\rm{Y}}\) are independent and uniformly distributed, the joint pdr ol \({\rm{X}}\) and \({\rm{Y}}\) is \(\begin{aligned}{*{20}{l}}{{\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y) = 1 \times 1 = 1,}}}&{{\rm{5£x£6,5£y£6}}}\\{{\rm{f(x,y) = 0}}}&{{\rm{ otherwise }}}\end{aligned}\)

Therefore,

\({\rm{f(x,y) = }}\left\{ {\begin{aligned}{*{20}{l}}{\rm{1}}&{{\rm{,5£x£6,5££6}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{aligned}} \right.\)

(b):

Random variables \({\rm{X}}\) and \({\rm{Y}}\) can take values between\({\rm{5 and 6}}\), which means that we first need to transfer time \({\rm{5:15}}\) uniformly on interval\({\rm{(5,6)}}\). Since \({\rm{1/4(0}}{\rm{.25)}}\) hours is \({\rm{15}}\) mins, \({\rm{5:15}}\)would be represented as\({\rm{5}}{\rm{.25}}\). Similarly, \({\rm{5:45}}\)would be represented as \({\rm{5}}{\rm{.75}}\) (\({\rm{3/4}}\)hours).

Therefore, the following is true

\(\begin{aligned}{\rm{P(5}}{\rm{.25£X£5}}{\rm{.75,5}}{\rm{.25£Y£5}}{\rm{.75)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(5}}{\rm{.25£X£5}}{\rm{.75) \times P(5}}{\rm{.25£Y£5}}{\rm{.75)}}\\{\rm{ = }}\int_{{\rm{5}}{\rm{.25}}}^{{\rm{5}}{\rm{.75}}} {\rm{1}} {\rm{dx \times }}\int_{{\rm{5}}{\rm{.25}}}^{{\rm{5}}{\rm{.75}}} {\rm{1}} {\rm{dy}}\\{\rm{ = 0}}{\rm{.5 \times 0}}{\rm{.5}}\\{\rm{ = 0}}{\rm{.25}}\end{aligned}\)

(1): We can utilize the multiplication property since the random variable is independent.

Property of Multiplication: Two events If and only if, A and B are independent.

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

(c):

We are given the event of interest in the hint. It makes sense because \({\rm{1/6}}\) hours is \({\rm{10}}\) minutes.

For every adequate set \({\rm{A}}\) the following holds

We have joint pdf, therefore, the following is true

(1): We can simply determine the area of a darkened region on the picture since random variables are equally distributed over a square. The area may be computed using the formula:

\(\begin{aligned}{\rm{1 - 2 \times P(\Delta ) = 1 - 2 \times a \times b \times }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{ = 1 - 2 \times }}\frac{{\rm{5}}}{{\rm{6}}}{\rm{ \times }}\frac{{\rm{5}}}{{\rm{6}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{ = 1 - }}\frac{{{\rm{25}}}}{{{\rm{36}}}}\\{\rm{ = }}\frac{{{\rm{11}}}}{{{\rm{36}}}}{\rm{,}}\end{aligned}\)

where \({\rm{a}}\) and \({\rm{b}}\) are the legs of the right triangle.