\begin{equation*}f(x,y)=\left\{\begin{array}{ll}xe^{-x(1+y)}&\quad x\geq 0\space and\space y\geq 0 \\ 0 &\quad otherwise\end{array}\right.\end{equation*}

**Find the probability that the lifetime****X****Â of the first component exceeds**Â**3.****Find the marginal probability density functions.****Find the probability that the lifespan of at most one component surpasses****5**

This problem aims to familiarize us with **probability** and **statistics.** The concepts required to solve this problem are **probability density functions, random variables,** and **marginal distribution functions.**

In probability, the **Probability Density Function** or **PDF** describes the probability function illustrating the **distribution** of a **continuous random variable** existing between a distinct range of **values.** Or we can say that the probability density function has the **probability** of values of the **continuous** random variable. The **formula** to find the **probability density function** is given:

\[P(a<X<b)=\int_{a}^{b}f(x)dx\]

## Expert Answer

**Part a:**

Let’s consider **two random variables** $X$ and $Y$ that predict the **lifespan** of the two **components** of the **minicomputer.**

The **joint probability** density function is given in the **statement:**

\begin{equation*}f(x,y)=\left\{\begin{array}{ll}xe^{-x(1+y)}&\quad x\geq 0\space and\space y\geq 0 \\ 0 &\quad otherwise\end{array}\right.\end{equation*}

The **required probability** does not **rely** on the values of $y$, so we will assume all the **potential** values of $Y$, and take the values from $3$ to $\infty$ for $X$ as the first **component surpasses** $3$.

Thus the **required probability** is:

\[P(x>3)=\int_{3}^{\infty}\int_{0}^{\infty} xe^{-x(1+y)} dydx\]

\[=\int_{3}^{\infty}([ -e^{-x(1+y)}]_{0}^{\infty}) dx\]

\[=\int_{3}^{\infty}e^x dx\]

\[=[\dfrac{-e^{-x}}{-1}]_{3}^{\infty}\]

\[P(x>3)\approx 0.05\]

So we get a **probability** of $0.05$ which **indicates** that there are only $5\%$ chances that the **lifespan** $X$ of the first **component** will **surpass** $3$.

**Part b:**

To find the **marginal probability density function** of $X$, we will **substitute** the provided **probability density function** and **integrate** it with respect to $y$:

\[f_x(x)=\int_{\infty}^{\infty}f(x,y)dy\space for -\infty\]

\[=\int_{0}^{\infty} xe^{-x(1+y)}dy\]

\[= [-e^{-x(1+y)}]_{0}^{\infty}\]

Now to find the **marginal probability density function** of $Y$, we will substitute the **provided** probability density function and **integrate** it with respect to $x$:

\[ f_y(y)=\int_{0}^{\infty}xe^{-x(1+y)}dx\]

\[=[\dfrac{xe^{-x(1+y)}}{-(1+y)}]_{0}^{\infty}-\int_{0}^{\infty} \dfrac{xe^{-x(1+y)}} {-(1+y)}dx\]

\[=[\dfrac{((y-1)x+1)e^{-yx-z}}{y^2+2y-1}]_{0}^{\infty}\]

\[=\dfrac{1}{(1+y)^2}\]

This represents the separate **probability **of occurrence of a **random variable** without assuming the occurrence of the other **variable.**

Now, to find whether the **two lifetimes** are **independent,** plug in the calculated **marginal PDF** and the **joint PDF** in the condition for **independence.**

\[f(x,y) = f_x(x)\times f_y(y)\]

\[xe^{-x(1+y)} \neq (e^{-x})(\dfrac{1}{(1+y)^2})\]

Since the **product** of **marginal PDF** is not equivalent to the given **joint** **PDF**, the two lifespans are **dependent.**

**Part c:**

The **probability** that the **lifespan** of at most one component **surpasses** $3$ is given by:

\[P(X>3\space or\space Y>3) =1- P(X,Y \leq 3)\]

\[=1-\int_{0}^{3}\int_{0}^{3} xe^{-x(1+y)} dydx\]

\[=1- \int_{0}^{3}([ -e^{-x(1+y)}]_{0}^{3}dx\]

\[=1-\int_{0}^{3}(( -e^{-4x}(e^{3x} -1))dx\]

Simplifying the **probability:**

\[P(X>3\space or\space Y>3)=1- [\dfrac{e^{-4x}}{4} – e -x]_{0}^{3}\]

\[=1-0.700\]

\[=0.3000\]

The **probability** indicates that there is only a $30\%$ chance that the **lifespan** of at most one **component** will **surpass** $3$.

## Numerical Result

**Part a:Â ** $P(x>3)\approx 0.05$

**Part b:** The two **lifespans** are **dependent.**

**Part c:** $30\%$ chance to **surpass** $3$.

## Example

If $X$ is a **continuous random variable** with **PDF:**

\begin{equation*}f(x)=\left\{\begin{array}{lll}x;&\quad 0<x<0 \\ 2-x;&\quad 1<x<2 \\ 0;&\quad x>2\end{array}\right.\end{equation*}

Then **find** $P(0.5<x<1.5)$.

\[P(0.5<x<1.5)=\int_{0.5}^{1.5}f(x)dx\]

**Splitting** the **integral:**

\[=\int_{0.5}^{1}f(x)dx+\int_{1}^{1.5}f(x)dx\]

**Substituting** the values:

\[=\int_{0.5}^{1}xdx+\int_{1}^{1.5}(2-x)dx\]

\[=[\dfrac{x^2}{2}]_{0.5}^{1}+[2x-\dfrac{x^2}{2}]_{1}^{1.5}\]

\[=\dfrac{3+15-12}{8} \]

\[=\dfrac{3}{4}\]