1.2.1 By Graph

Figure for Question 1a

Graph the line \(X< \frac{1}{2}-Y\) on a graph, we have the shaded area within the unit square \[=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} =\frac{1}{8}\]

1.2.2 By Integration

\[P(B+C<1/2)\] \[=P(0<x<\frac{1}{2}, 0<y<\frac{1}{2}-x)\] \[=\int_{0}^{1/2} \int_{0}^{1/2-x} f(x,y) dydx\] \[=\int_{0}^{1/2} \int_{0}^{1/2-x} 1dydx\] \[=\int_{0}^{1/2} [\frac{1}{2}-x]dx\] \[=\left[ \frac{x}{2}-\frac{x^{2}}{2} \right]^{1/2}_{0}\] \[=\frac{1}{4}-\frac{1}{8}\] \[=\frac{1}{8}\]

1.2.3 By Code

integrate((function(x){
  integrate(function(y) {0*y+1},0,0.5-x)$value}) %>% 
    Vectorize(),0,0.5)

## 0.125 with absolute error < 1.4e-15

1.3.1 By Graph

Figure for Question 1bi

Figure for Question 1bii

As the graph is a parabola, we cannot solve it by simple multiplication.

1.3.2 By Integration

\[P(BC < \frac{1}{2})\] \[=P(xy < \frac{1}{2})\] \[=P(y < \frac{1}{2x})\] \[=\int_{0}^{1/2} \int_{0}^{1} 1 dydx + \int_{1/2}^{1} \int_{0}^{\frac{1}{2x}} 1 dydx \]

\[=\frac{1}{2} + \int_{1/2}^{1} \frac{1}{2x} dx\]

\[=\frac{1}{2} + \frac{1}{2}\cdot \left [ ln(x) \right ]^{1}_{1/2} \] \[=\frac{1}{2} - \frac{1}{2} \cdot (ln(\frac{1}{2})) \] \[\approx 0.84657359\]

1.3.3 By Code

integrate((function(x){
  integrate(function(y) {0*y+1},0,1)$value}) %>% 
    Vectorize(),0,0.5)$value +
integrate((function(x){
  integrate(function(y) {0*y+1},0,1/(2*x))$value}) %>% 
    Vectorize(),0.5,1)$value

## [1] 0.8465736

1.4.1 By Graph

Figure for Question 1c

Graph the line \(|x-y|<\frac{1}{2}\) on a graph, calculate the shaded area within the unit square as sum of two symmetric trapezoids \[=(\frac{1}{2}+1)\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot2=\frac{3}{4}\]

1.4.2 By Integration

\[P(\left | B-C \right | < \frac{1}{2})\] \[=P(-\frac{1}{2} < x-y < \frac{1}{2})\] \[=P((y<x+\frac{1}{2}) and (y>x-\frac{1}{2}))\] \[=\int_{0}^{1/2} \int_{0}^{x+1/2} 1 dydx + \int_{1/2}^{1} \int_{x-1/2}^{1} 1 dydx\] \[=\int_{0}^{1/2}(x+\frac{1}{2})dx + \int_{1/2}^{1}(\frac{3}{2}-x)dx\] \[=\left [ \frac{x^{2}}{2}+\frac{x}{2} \right ]^{1/2}_{0} + \left [ \frac{3x}{2} - \frac{x^{2}}{2} \right ]^{1}_{1/2}\] \[=(\frac{1}{8}+\frac{1}{4})-0+(\frac{3}{2}-\frac{1}{2})-(\frac{3}{4}-\frac{1}{8})\] \[=\frac{3}{4}\]

1.4.3 By Code

integrate((function(x){
  integrate(function(y) {0*y+1},0,x+0.5)$value}) %>% 
    Vectorize(),0,0.5)$value +
integrate((function(x){
  integrate(function(y) {0*y+1},x-0.5,1)$value}) %>% 
    Vectorize(),0.5,1)$value

## [1] 0.75

1.5.1 By Graph

Figure for Question 1d

Graph the line \(max\left\{B,C\right\}<\frac{1}{2}\) on a graph, calculate the shaded area within the unit square \[=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\]

1.5.2 By Integration

\[P(max\left\{B,C\right\}<\frac{1}{2})\] \[=P(max\left\{x,y\right\}<\frac{1}{2})\] \[=P((0<x<\frac{1}{2}) and (0<y<\frac{1}{2}))\] \[\int_{0}^{1/2}\int_{0}^{1/2} 1 \;dy\,dx\] \[=\frac{1}{2} \cdot \frac{1}{2}\] \[=\frac{1}{4}\]

1.5.3 By Code

integrate((function(x){
  integrate(function(y) {0*y+1},0,0.5)$value}) %>% 
    Vectorize(),0,0.5)$value

## [1] 0.25

1.6.1 By Graph

Figure for Question 1e

Graph the line \(min\left\{B,C\right\}<\frac{1}{2}\) on a graph, calculate the shaded area within the unit square \[=1-\frac{1}{2}\cdot\frac{1}{2}=\frac{3}{4}\]

1.6.2 By Integration

\[P(min\left\{B,C\right\}<\frac{1}{2})\] \[=P(min\left\{x,y\right\}<\frac{1}{2})\] \[=P((x<\frac{1}{2})or(y<\frac{1}{2}))\] \[=\int_{0}^{1/2}\int_{0}^{1} 1dydx + \int_{1/2}^{1}\int_{0}^{1/2}1dydx\] \[=\frac{1}{2} + \frac{1}{4}\] \[=\frac{3}{4}\]

1.6.3 By Code

integrate((function(x){
  integrate(function(y) {0*y+1},0,1)$value}) %>% 
    Vectorize(),0,0.5)$value +
integrate((function(x){
  integrate(function(y) {0*y+1},0,0.5)$value}) %>% 
    Vectorize(),0.5,1)$value

## [1] 0.75