Graded Assignment 6

1. Uniform Distribution on a Circle

Question: Let $(X, Y) \sim \text{Uniform}(D)$, where $D = {(x, y) : (x - 3)^2 + (y - 3)^2 \le 1}$. Calculate $P(X > Y)$. Enter your answer correct to two decimals accuracy.

Concept: 🗺️ Geometric Probability

When a point $(X, Y)$ is chosen uniformly from a region $D$, the probability of an event $A$ (which corresponds to a sub-region) is the ratio of the areas:

Detailed Solution:

1. Identify the Region $D$: The equation $(x - 3)^2 + (y - 3)^2 \le 1$ describes a circle $D$ centered at the point $(3, 3)$ with a radius $r = 1$.
2. Calculate $\text{Area}(D)$: The area of this circle is $\text{Area}(D) = \pi r^2 = \pi(1)^2 = \pi$.
3. Identify the Event $A$: The event is $P(X > Y)$, which can be rewritten as $Y < X$. This corresponds to the region $A$ of the $xy$-plane that lies below the line $y = x$.
4. Find the $\text{Area}(A \cap D)$: We need the area of the part of the circle $D$ that is also below the line $y = x$.
5. Analyze the Geometry: Notice that the center of the circle is $(3, 3)$. This point lies exactly on the line $y = x$. Any line that passes through the center of a circle divides the circle into two equal semicircles. 6. Therefore, the line $y = x$ cuts the circle $D$ into two equal halves. The region $Y < X$ (below the line) represents exactly half of the circle’s area.
6. $\text{Area}(A \cap D) = \frac{1}{2} \times \text{Area}(D) = \frac{\pi}{2}$.
7. Calculate the Probability: $$P(X > Y) = \frac{\text{Area}(A \cap D)}{\text{Area}(D)} = \frac{\pi / 2}{\pi} = \frac{1}{2} = 0.5$$

Answer: 0.50

2. Marginal Density from a Uniform Distribution

Question: Let $(X, Y) \sim \text{Uniform}(D)$, where $D = {(x, y) : y \le 2x, 0 < x < 1, 0 < y < 2} \cup {[1, 2] \times [0, 2]}$. Find the marginal density of $X$, $f_X(x)$.

Concept: 🧩 Marginal Density

1. Joint PDF for Uniform: The joint PDF is $f_{XY}(x, y) = \frac{1}{\text{Area}(D)}$ if $(x, y)$ is in $D$, and 0 otherwise.
2. Marginal PDF: The marginal PDF $f_X(x)$ is found by “integrating out” $y$ from the joint PDF. It tells us the probability distribution of $X$ regardless of $Y$’s value. $$f_X(x) = \int_{-\infty}^{\infty} f_{XY}(x, y) \,dy$$

Detailed Solution:

1. Sketch and Find $\text{Area}(D)$:
   • Part 1 ($D_1$): $0 < x < 1$, $0 < y < 2$, and $y \le 2x$. This forms a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,2)$. The area is $\text{Area}(D_1) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1$.
   • Part 2 ($D_2$): $[1, 2] \times [0, 2]$. This is a rectangle with width $(2-1)=1$ and height $(2-0)=2$. The area is $\text{Area}(D_2) = 1 \times 2 = 2$.
   • Total Area: $\text{Area}(D) = \text{Area}(D_1) + \text{Area}(D_2) = 1 + 2 = 3$.
2. Find the Joint PDF: $f_{XY}(x, y) = \frac{1}{\text{Area}(D)} = \frac{1}{3}$, for any $(x, y) \in D$.
3. Find the Marginal $f_X(x)$: We must do this in parts, based on the value of $x$.
   • Case 1: $0 \le x \le 1$ (corresponding to $D_1$) In this interval, for a given $x$, $y$ ranges from $0$ up to $2x$. $$f_X(x) = \int_0^{2x} f_{XY}(x, y) \,dy = \int_0^{2x} \frac{1}{3} \,dy = \frac{1}{3} [y]_0^{2x} = \frac{1}{3}(2x - 0) = \frac{2x}{3}$$
   • Case 2: $1 \le x \le 2$ (corresponding to $D_2$) In this interval, for a given $x$, $y$ ranges from $0$ up to $2$. $$f_X(x) = \int_0^2 f_{XY}(x, y) \,dy = \int_0^2 \frac{1}{3} \,dy = \frac{1}{3} [y]_0^2 = \frac{1}{3}(2 - 0) = \frac{2}{3}$$
   • Case 3: Otherwise $f_X(x) = 0$.
4. Combine the pieces: $f_X(x) = \begin{cases} \frac{2x}{3} & 0 \le x \le 1 \ \frac{2}{3} & 1 < x \le 2 \ 0 & \text{otherwise} \end{cases}$

Answer: The correct choice is the third option: $f_X(x) = \begin{cases} \frac{2x}{3} & 0 \le x \le 1 \ \frac{2}{3} & 1 \le x \le 2 \ 0 & \text{otherwise} \end{cases}$

3. Probability with Independent Exponential Variables

Question: The joint pdf of two continuous random variables $X$ and $Y$ is given by $f_{XY}(x, y) = \begin{cases} ke^{-(x+y)} & x \ge 0, y \ge 0 \ 0 & \text{otherwise} \end{cases}$. Find the value of $P(X \ge 3, Y \le 6)$.

Concept: 🔗 Independence

1. Normalization: The constant $k$ is found by ensuring the total probability is 1: $\iint f_{XY}(x, y) ,dx ,dy = 1$.
2. Independence: Two variables $X$ and $Y$ are independent if their joint PDF $f_{XY}(x, y)$ can be factored into the product of their marginal PDFs: $f_{XY}(x, y) = f_X(x) f_Y(y)$.
3. Probability of Independent Events: If $X$ and $Y$ are independent, $P(A \cap B) = P(A) \times P(B)$. Therefore, $P(X \ge 3, Y \le 6) = P(X \ge 3) \times P(Y \le 6)$.

Detailed Solution:

1. Find $k$: $$\int_0^\infty \int_0^\infty ke^{-(x+y)} \,dx \,dy = k \left( \int_0^\infty e^{-x} \,dx \right) \left( \int_0^\infty e^{-y} \,dy \right) = 1$$ The integral $\int_0^\infty e^{-x} ,dx = [-e^{-x}]_0^\infty = (0) - (-1) = 1$. So, $k \times (1) \times (1) = 1$, which means $k = 1$.
2. Analyze the PDF: The joint PDF is $f_{XY}(x, y) = e^{-(x+y)} = e^{-x} e^{-y}$. This can be factored into $f_X(x) = e^{-x}$ (for $x \ge 0$) and $f_Y(y) = e^{-y}$ (for $y \ge 0$). This confirms $X$ and $Y$ are independent.
3. Calculate $P(X \ge 3)$: $$P(X \ge 3) = \int_3^\infty f_X(x) \,dx = \int_3^\infty e^{-x} \,dx = [-e^{-x}]_3^\infty = (0) - (-e^{-3}) = e^{-3}$$
4. Calculate $P(Y \le 6)$: $$P(Y \le 6) = \int_0^6 f_Y(y) \,dy = \int_0^6 e^{-y} \,dy = [-e^{-y}]_0^6 = (-e^{-6}) - (-e^0) = 1 - e^{-6}$$
5. Calculate the Joint Probability: $$P(X \ge 3, Y \le 6) = P(X \ge 3) \times P(Y \le 6) = e^{-3} (1 - e^{-6})$$

Answer: The correct choice is the fourth option: $(1 - e^{-6})e^{-3}$

4. Conditional Probability

Question: The joint pdf of two random variables $X$ and $Y$ is given by $f_{XY}(x, y) = \begin{cases} \frac{1}{8}(x+y) & 0 < x < 2, 0 < y < 2 \ 0 & \text{otherwise} \end{cases}$. Find the value of $P(\frac{2}{9} \le Y \le 2 \mid X = \frac{1}{9})$.

Concept: 📊 Conditional Density

To find the probability of $Y$ given a specific value of $X$, we must first find the conditional probability density function, $f_{Y|X}(y|x)$.

1. Find the marginal PDF of $X$: $f_X(x) = \int f_{XY}(x, y) ,dy$
2. Find the conditional PDF: $f_{Y|X}(y|x) = \frac{f_{XY}(x, y)}{f_X(x)}$
3. Integrate the conditional PDF over the desired range of $y$: $P(a \le Y \le b \mid X = x_0) = \int_a^b f_{Y|X}(y|x_0) ,dy$

Detailed Solution:

1. Find $f_X(x)$ (the marginal density of $X$): We integrate the joint PDF with respect to $y$ over its domain $[0, 2]$. $$f_X(x) = \int_0^2 \frac{1}{8}(x+y) \,dy = \frac{1}{8} \left[ xy + \frac{y^2}{2} \right]_0^2$$ $$f_X(x) = \frac{1}{8} \left( (x(2) + \frac{2^2}{2}) - (0) \right) = \frac{1}{8} (2x + 2) = \frac{x+1}{4}$$ This is valid for $0 < x < 2$.
2. Find $f_{Y|X}(y|x)$ (the conditional density of $Y$ given $X$): $$f_{Y|X}(y|x) = \frac{f_{XY}(x, y)}{f_X(x)} = \frac{\frac{1}{8}(x+y)}{\frac{1}{4}(x+1)} = \frac{1}{2} \frac{x+y}{x+1}$$ This is the formula for the PDF of $Y$, given any value $x$.
3. Evaluate at $x = 1/9$: We need the specific PDF for $X = 1/9$. $$f_{Y|X}(y|\frac{1}{9}) = \frac{1}{2} \frac{(1/9) + y}{(1/9) + 1} = \frac{1}{2} \frac{(1/9) + y}{10/9} = \frac{1}{2} \frac{9}{10} (y + \frac{1}{9}) = \frac{9}{20}(y + \frac{1}{9})$$
4. Calculate the Probability: We integrate this conditional PDF from $y = 2/9$ to $y = 2$. $$P(\frac{2}{9} \le Y \le 2 \mid X = \frac{1}{9}) = \int_{2/9}^2 \frac{9}{20}(y + \frac{1}{9}) \,dy$$ $$= \frac{9}{20} \left[ \frac{y^2}{2} + \frac{y}{9} \right]_{2/9}^2$$ $$= \frac{9}{20} \left[ \left( \frac{2^2}{2} + \frac{2}{9} \right) - \left( \frac{(2/9)^2}{2} + \frac{2/9}{9} \right) \right]$$ $$= \frac{9}{20} \left[ \left( 2 + \frac{2}{9} \right) - \left( \frac{4/81}{2} + \frac{2}{81} \right) \right]$$ $$= \frac{9}{20} \left[ \left( \frac{20}{9} \right) - \left( \frac{2}{81} + \frac{2}{81} \right) \right]$$ $$= \frac{9}{20} \left[ \frac{20}{9} - \frac{4}{81} \right]$$ $$= \left( \frac{9}{20} \frac{20}{9} \right) - \left( \frac{9}{20} \frac{4}{81} \right) = 1 - \frac{1}{20} \frac{4}{9} = 1 - \frac{1}{5 \times 9} = 1 - \frac{1}{45} = \frac{44}{45}$$
5. Convert to Decimal: $44 \div 45 \approx 0.9777…$

Answer: 0.98

5. Probability over a Region

Question: The joint density function of two random variables $X$ and $Y$ is $f_{XY}(x, y) = \begin{cases} \frac{2}{3}(x+1) & 0 < x < 1, 0 < y < 1 \ 0 & \text{otherwise} \end{cases}$. Find the value of $P(X > Y)$.

Concept: ∬ Integrating over a Region

To find the probability of an event (like $X > Y$), you must integrate the joint PDF $f_{XY}(x, y)$ over the region $A$ defined by that event, within the bounds of the PDF’s domain $D$.

Detailed Solution:

1. Identify the Domain $D$: The domain is the unit square, $[0, 1] \times [0, 1]$.
2. Identify the Event Region $A$: The event is $X > Y$, or $y < x$.
3. Identify the Integration Region $A \cap D$: We need the part of the unit square that is also below the line $y = x$. This is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. 4. Set up the Double Integral: We will integrate $f_{XY}(x, y) = \frac{2}{3}(x+1)$ over this triangle. It’s easiest to set the outer integral for $x$ and the inner for $y$.
   • $x$ ranges from $0$ to $1$.
   • For any given $x$, $y$ ranges from $0$ up to the line $y=x$. $$P(X > Y) = \int_0^1 \left( \int_0^x \frac{2}{3}(x+1) \,dy \right) \,dx$$
4. Solve the Inner Integral (with respect to $y$): The term $\frac{2}{3}(x+1)$ is treated as a constant. $$\int_0^x \frac{2}{3}(x+1) \,dy = \frac{2}{3}(x+1) [y]_0^x = \frac{2}{3}(x+1)(x - 0) = \frac{2}{3}(x^2 + x)$$
5. Solve the Outer Integral (with respect to $x$): $$\int_0^1 \frac{2}{3}(x^2 + x) \,dx = \frac{2}{3} \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_0^1$$ $$= \frac{2}{3} \left[ \left( \frac{1^3}{3} + \frac{1^2}{2} \right) - (0) \right] = \frac{2}{3} \left( \frac{1}{3} + \frac{1}{2} \right)$$ $$= \frac{2}{3} \left( \frac{2}{6} + \frac{3}{6} \right) = \frac{2}{3} \left( \frac{5}{6} \right) = \frac{10}{18} = \frac{5}{9}$$
6. Convert to Decimal: $5 \div 9 = 0.5555…$

Answer: 0.56

6. Finding the Normalization Constant

Question: The joint density function of two random variables $X$ and $Y$ is given by $f(x, y) = \begin{cases} k(x+2y) & \text{for } 0 < x < 1, 1 < y < 2 \ 0 & \text{otherwise} \end{cases}$. Find the value of $k$.

Concept: ⚖️ Normalization

A probability density function (PDF) must “contain” a total probability of 1. This means the integral of the PDF over its entire domain must equal 1.

Detailed Solution:

1. Set up the Integral: We set the double integral of the PDF over its given domain equal to 1. $$\int_1^2 \left( \int_0^1 k(x+2y) \,dx \right) \,dy = 1$$
2. Solve the Inner Integral (with respect to $x$): $$k \int_1^2 \left[ \frac{x^2}{2} + 2yx \right]_0^1 \,dy$$ $$= k \int_1^2 \left( (\frac{1^2}{2} + 2y(1)) - (0) \right) \,dy = k \int_1^2 (\frac{1}{2} + 2y) \,dy$$
3. Solve the Outer Integral (with respect to $y$): $$k \left[ \frac{1}{2}y + y^2 \right]_1^2 = 1$$ $$k \left[ (\frac{1}{2}(2) + 2^2) - (\frac{1}{2}(1) + 1^2) \right] = 1$$ $$k \left[ (1 + 4) - (\frac{1}{2} + 1) \right] = 1$$ $$k \left[ 5 - \frac{3}{2} \right] = 1$$ $$k \left[ \frac{10}{2} - \frac{3}{2} \right] = 1$$ $$k \left( \frac{7}{2} \right) = 1$$
4. Solve for $k$: $$k = \frac{2}{7}$$
5. Convert to Decimal: $2 \div 7 \approx 0.285714…$

Answer: 0.286

7. Conditional Probability

Question: Let $X$ and $Y$ be two continuous random variables with joint PDF $f_{XY}(x, y) = \begin{cases} xe^{-x(y+1)} & x > 0, y > 0 \ 0 & \text{otherwise} \end{cases}$. Find the value of $P(Y > 2 \mid X > 1)$.

Concept: 📋 Conditional Probability Formula

The definition of conditional probability is:

In this case, $A = {Y > 2}$ and $B = {X > 1}$. So, we need to find:

Detailed Solution:

1. Calculate the numerator, $P(Y > 2, X > 1)$: We integrate the joint PDF over the region $x > 1$ and $y > 2$.

   $$\int_1^\infty \left( \int_2^\infty xe^{-x(y+1)} \,dy \right) \,dx = \int_1^\infty \left( \int_2^\infty xe^{-xy}e^{-x} \,dy \right) \,dx$$

   Let’s solve the inner integral (w.r.t $y$) first:

   $$\int_2^\infty xe^{-x}e^{-xy} \,dy = xe^{-x} \left[ \frac{e^{-xy}}{-x} \right]_2^\infty$$$$= xe^{-x} \left( (0) - (\frac{e^{-x(2)}}{-x}) \right) = xe^{-x} \left( \frac{e^{-2x}}{x} \right) = e^{-x}e^{-2x} = e^{-3x}$$

   Now solve the outer integral (w.r.t $x$):

   $$\int_1^\infty e^{-3x} \,dx = \left[ \frac{e^{-3x}}{-3} \right]_1^\infty = (0) - (\frac{e^{-3(1)}}{-3}) = \frac{e^{-3}}{3}$$

   So, $P(Y > 2, X > 1) = \frac{e^{-3}}{3}$.

2. Calculate the denominator, $P(X > 1)$: First, we need the marginal PDF $f_X(x)$ by integrating $f_{XY}$ w.r.t $y$ over its full domain $(0, \infty)$.

   $$f_X(x) = \int_0^\infty xe^{-x(y+1)} \,dy = \int_0^\infty xe^{-xy}e^{-x} \,dy$$$$= xe^{-x} \int_0^\infty e^{-xy} \,dy = xe^{-x} \left[ \frac{e^{-xy}}{-x} \right]_0^\infty$$$$= xe^{-x} \left( (0) - (\frac{e^0}{-x}) \right) = xe^{-x} \left( \frac{1}{x} \right) = e^{-x}$$

   (So, $X$ is an exponential variable with $\lambda=1$). Now, find $P(X > 1)$:

   $$P(X > 1) = \int_1^\infty f_X(x) \,dx = \int_1^\infty e^{-x} \,dx = [-e^{-x}]_1^\infty = (0) - (-e^{-1}) = e^{-1}$$

3. Calculate the Final Probability:

   $$P(Y > 2 \mid X > 1) = \frac{P(Y > 2, X > 1)}{P(X > 1)} = \frac{e^{-3} / 3}{e^{-1}} = \frac{1}{3} e^{-3} e^1 = \frac{1}{3}e^{-2}$$

Answer: The correct choice is the second option: $\frac{1}{3}e^{-2}$

8. Probability over a Triangular Region

Question: At a particular bakery, … $f_{XY}(x, y) = \begin{cases} cxy & \text{if } 0 \le y \le x \le 2 \ 0 & \text{otherwise} \end{cases}$. Find the probability that the amount of dough used ($Y$) … is less than half the amount prepared ($X$).

Concept: 1. Normalization, 2. Probability Calculation This is a two-step problem.

1. Find $c$: Use the normalization rule $\iint f_{XY}(x, y) ,dA = 1$ over the domain $D$.
2. Find Probability: The question asks for $P(Y < X/2)$. We must integrate the now-known PDF, $f_{XY}(x, y)$, over the event region $A = {Y < X/2}$ (intersected with $D$).

Detailed Solution:

1. Identify Domain $D$: The domain $D$ is $0 \le y \le x \le 2$. This is a triangle with vertices $(0,0)$, $(2,0)$, and $(2,2)$.

2. Find $c$:

   $$\int_0^2 \left( \int_0^x cxy \,dy \right) \,dx = 1$$

   Inner integral (w.r.t $y$): $c \int_0^2 x \left[ \frac{y^2}{2} \right]_0^x ,dx = c \int_0^2 x (\frac{x^2}{2}) ,dx = c \int_0^2 \frac{x^3}{2} ,dx$ Outer integral (w.r.t $x$): $\frac{c}{2} \left[ \frac{x^4}{4} \right]0^2 = \frac{c}{8} (2^4 - 0) = \frac{c}{8} (16) = 2c$. Since $2c = 1$, we have $c = \frac{1}{2}$. The PDF is $f{XY}(x, y) = \frac{1}{2}xy$.

3. Find $P(Y < X/2)$: We must integrate $f_{XY}(x, y) = \frac{1}{2}xy$ over the region $A \cap D$.

   • $D$ is $0 \le y \le x \le 2$.
   • $A$ is $y < x/2$. The combined region is $0 \le y \le x/2$ and $0 \le x \le 2$. (The condition $y \le x/2$ automatically satisfies $y \le x$, since $x/2 < x$ for $x>0$).

4. Set up the Integral:

   $$P(Y < X/2) = \int_0^2 \left( \int_0^{x/2} \frac{1}{2}xy \,dy \right) \,dx$$

5. Solve the Inner Integral (w.r.t $y$):

   $$\frac{1}{2} \int_0^2 x \left[ \frac{y^2}{2} \right]_0^{x/2} \,dx = \frac{1}{2} \int_0^2 x \left( \frac{(x/2)^2}{2} - 0 \right) \,dx$$$$= \frac{1}{2} \int_0^2 x \left( \frac{x^2/4}{2} \right) \,dx = \frac{1}{2} \int_0^2 \frac{x^3}{8} \,dx = \frac{1}{16} \int_0^2 x^3 \,dx$$

6. Solve the Outer Integral (w.r.t $x$):

   $$\frac{1}{16} \left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{64} (2^4 - 0) = \frac{16}{64} = \frac{1}{4}$$

Answer: 0.25

9. Conditional Probability (Independent Variables)

Question: … The joint probability density function of $X$ and $Y$ is given by $f_{XY}(x, y) = \begin{cases} \frac{x^3y}{k} & 0 < x < 1, 0 < y < 2 \ 0 & \text{otherwise} \end{cases}$. If a patient is known to have received a 0.5 ml dose ($X=0.5$), what is the probability that the duration of relief lies between 0.5 and 1.5 hours ($0.5 \le Y \le 1.5$)?

Concept: Conditional Probability & Independence

This problem asks for $P(0.5 \le Y \le 1.5 \mid X = 0.5)$. We can solve this using the conditional density $f_{Y|X}(y|x)$, just like in problem 4. However, we should first check if $X$ and $Y$ are independent.

Detailed Solution:

1. Find $k$: $$\int_0^1 \int_0^2 \frac{x^3y}{k} \,dy \,dx = 1$$ $$\frac{1}{k} \left( \int_0^1 x^3 \,dx \right) \left( \int_0^2 y \,dy \right) = 1$$ $$\frac{1}{k} \left[ \frac{x^4}{4} \right]_0^1 \left[ \frac{y^2}{2} \right]_0^2 = 1$$ $$\frac{1}{k} \left( \frac{1}{4} \right) \left( \frac{4}{2} \right) = 1 \implies \frac{1}{k} \left( \frac{1}{2} \right) = 1 \implies k = \frac{1}{2}$$
2. Analyze the PDF: The joint PDF is $f_{XY}(x, y) = 2x^3y$. The domain is a rectangle $[0, 1] \times [0, 2]$. The function $2x^3y$ can be factored into a function of $x$ and a function of $y$: $f_{XY}(x, y) = (4x^3) \times (\frac{y}{2})$. (We can find these marginals: $f_X(x) = \int_0^2 2x^3y ,dy = 2x^3 [\frac{y^2}{2}]_0^2 = 4x^3$. And $f_Y(y) = \int_0^1 2x^3y ,dx = 2y [\frac{x^4}{4}]0^1 = \frac{y}{2}$. Since $f_X(x)f_Y(y) = (4x^3)(\frac{y}{2}) = 2x^3y = f{XY}(x, y)$, the variables are independent.)
3. Calculate the Probability: Because $X$ and $Y$ are independent, the condition $X=0.5$ provides no information about $Y$. $$P(0.5 \le Y \le 1.5 \mid X = 0.5) = P(0.5 \le Y \le 1.5)$$ We can find this using the marginal PDF for $Y$, which we found to be $f_Y(y) = \frac{y}{2}$ for $0 < y < 2$. $$P(0.5 \le Y \le 1.5) = \int_{0.5}^{1.5} f_Y(y) \,dy = \int_{0.5}^{1.5} \frac{y}{2} \,dy$$ $$= \frac{1}{2} \left[ \frac{y^2}{2} \right]_{0.5}^{1.5} = \frac{1}{4} [y^2]_{0.5}^{1.5}$$ $$= \frac{1}{4} (1.5^2 - 0.5^2) = \frac{1}{4} (2.25 - 0.25) = \frac{1}{4} (2) = \frac{1}{2}$$

Answer: 0.5

10. Uniform Distribution on a Rectangle

Question: Let $(X, Y) \sim \text{Uniform}(D)$, where $D := [0, 5] \times [0, 2]$. Find $P(X + Y < 4)$. Enter the answer correct to one decimal place.

Concept: 🗺️ Geometric Probability

Just like in problem 1, the probability is the ratio of the areas.

Detailed Solution:

1. Identify the Region $D$: This is a rectangle with vertices $(0,0)$, $(5,0)$, $(5,2)$, and $(0,2)$.
2. Calculate $\text{Area}(D)$: $\text{Area}(D) = \text{width} \times \text{height} = 5 \times 2 = 10$.
3. Identify the Event $A$: The event is $X + Y < 4$, which can be rewritten as $Y < 4 - X$. This is the region below the line $y = 4 - x$.
4. Find $\text{Area}(A \cap D)$: We need to find the area of the rectangle $D$ that is also below the line $y = 4 - x$.
   • Let’s sketch the line $y = 4 - x$. It hits the $x$-axis at $(4,0)$ and the $y$-axis at $(0,4)$.
   • Let’s see where this line intersects the boundaries of our rectangle $D$:
     • Boundary $y=0$: The line hits at $x=4$. (The point $(4,0)$ is on the edge of $D$).
     • Boundary $y=2$: The line hits at $2 = 4 - x \implies x = 2$. (The point $(2,2)$ is on the edge of $D$).
   • The region $A \cap D$ is the part of the rectangle bounded by $x=0$, $y=0$ and the line $y = 4-x$ (and the top boundary $y=2$).
   • x[0,2] with the line y=4-x, showing the area of interest]
   • The resulting shape is a polygon with vertices $(0,0)$, $(4,0)$, $(2,2)$, and $(0,2)$.
5. Calculate the Area of the Polygon: We can split this polygon into a rectangle and a triangle.
   • Rectangle: From $x=0$ to $x=2$, the region is bounded by $y=0$ and $y=2$. $\text{Area} = 2 \times 2 = 4$.
   • Triangle: From $x=2$ to $x=4$, the region is a triangle with vertices $(2,0)$, $(4,0)$, and $(2,2)$. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4-2) \times 2 = 2$.
   • Total Area: $\text{Area}(A \cap D) = 4 + 2 = 6$.
   • (Note: For $x > 4$, the line $y = 4-x$ is below $y=0$, so none of that area is counted).
6. Calculate the Probability: $$P(X + Y < 4) = \frac{\text{Area}(A \cap D)}{\text{Area}(D)} = \frac{6}{10} = 0.6$$

Answer: 0.6