# Two components of a minicomputer have the following joint PDF for their useful lifetimes X and Y:

1. Find the probability that the lifetime X of the first component exceeds 3.
2. Find the marginal probability density functions.
3. 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:

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

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:

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:  $P(x>3)\approx 0.05$

Part b: The two lifespans are dependent.

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

## Example

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

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:

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

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

$=\dfrac{3+15-12}{8}$

$=\dfrac{3}{4}$