**Probability with Continuous Random Variables: Example Problem**---**Problem Statement:**Let $ X $ and $ Y $ be continuous random variables with the following joint probability density function (pdf):\[ f(x,y) = \begin{cases} \frac{1}{2}x^3 y, & \text{if } 0 < x < 2 \text{ and } 0 < y < 1 \\0, & \text{otherwise}\end{cases}\]Find the probability that $ X $ is less than $ 1 $ and $ Y $ is less than $ \frac{1}{2} $.**Solution Approach:**To find this probability, we need to integrate the joint probability density function $ f(x,y) $ over the appropriate ranges for $ X $ and $ Y $:\[P(X < 1, Y < \frac{1}{2}) = \int_0^{1} \int_0^{\frac{1}{2}} f(x,y) \, dy \, dx.\]**Calculation Steps:**1. **Define the integration limits based on the conditions**: - For $ X $: from $ 0 $ to $ 1 $ - For $ Y $: from $ 0 $ to $ \frac{1}{2} $2. **Substitute the given pdf into the integral**:\[P(X < 1, Y < \frac{1}{2}) = \int_0^{1} \int_0^{\frac{1}{2}} \frac{1}{2} x^3 y \, dy \, dx\]3. **Integrate with respect to $ y $**:\[\int_0^{\frac{1}{2}} \frac{1}{2} x^3 y \, dy = \frac{1}{2} x^3 \int_0^{\frac{1}{2}} y \, dy = \frac{1}{2} x^3 \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{2} x^3 \left( \frac{\left( \frac{1}{2} \right)^2}{2} -

Joint Probability Distribution: Let X and Y be two continuous random variables defined over the…

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Let X and Y be continuous random variable with the following joint probability density function 1 x³y, if 0

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

**Probability with Continuous Random Variables: Example Problem**

---

**Problem Statement:**

Let $ X $ and $ Y $ be continuous random variables with the following joint probability density function (pdf):

\[
f(x,y) =
\begin{cases}
\frac{1}{2}x^3 y, & \text{if } 0 < x < 2 \text{ and } 0 < y < 1 \\
0, & \text{otherwise}
\end{cases}
\]

Find the probability that $ X $ is less than $ 1 $ and $ Y $ is less than $ \frac{1}{2} $.

**Solution Approach:**

To find this probability, we need to integrate the joint probability density function $ f(x,y) $ over the appropriate ranges for $ X $ and $ Y $:
\[
P(X < 1, Y < \frac{1}{2}) = \int_0^{1} \int_0^{\frac{1}{2}} f(x,y) \, dy \, dx.
\]

**Calculation Steps:**

1. **Define the integration limits based on the conditions**:
- For $ X $: from $ 0 $ to $ 1 $
- For $ Y $: from $ 0 $ to $ \frac{1}{2} $

2. **Substitute the given pdf into the integral**:
\[
P(X < 1, Y < \frac{1}{2}) = \int_0^{1} \int_0^{\frac{1}{2}} \frac{1}{2} x^3 y \, dy \, dx
\]

3. **Integrate with respect to $ y $**:
\[
\int_0^{\frac{1}{2}} \frac{1}{2} x^3 y \, dy = \frac{1}{2} x^3 \int_0^{\frac{1}{2}} y \, dy = \frac{1}{2} x^3 \left[ \frac{y^2}{2} \right]_0^{\frac{1}{2}} = \frac{1}{2} x^3 \left( \frac{\left( \frac{1}{2} \right)^2}{2} -

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