Question

Let the random variables \(X\) and \(Y\) denote the time to failure of component A and component B, respectively, of an instrument. The joint probability density function of \((X,Y)\) is given by

• A. \(\dfrac{e\sqrt{e}-1}{e\sqrt{e}}\)
• B. \(\dfrac{e\sqrt{e}}{e\sqrt{e}+1}\)
• C. \(\dfrac{e\sqrt{e}-1}{e\sqrt{e}+1}\)
• D. \(\dfrac{1}{e\sqrt{e}}\)

Hint:

For probability questions involving joint density, first translate the event into inequalities and then combine them with the support of the density.

Answer 1

The Correct Option is A

Step 1: State the event.
B fails before A means $Y<X$, and the density is positive for $5<x<10$, $y>x-3$.

Step 2: Set the limits.
For each $x$, $y$ runs from $x-3$ up to $x$, with $x$ from $5$ to $10$.

Step 3: Do the inner integral.
Let $u=y+3-x$, so $u$ goes $0$ to $3$. Then $\int_{x-3}^x \frac1{10}e^{-(y+3-x)/2}dy=\frac1{10}\int_0^3 e^{-u/2}du=\frac15\left(1-e^{-3/2}\right)$.

Step 4: Do the outer integral.
The integrand is constant in $x$ over a length $5$ interval, so the answer is $1-e^{-3/2}$.

Step 5: Tidy up.
Since $e^{3/2}=e\sqrt e$, this is $\dfrac{e\sqrt e-1}{e\sqrt e}$, option (A).
\[ \boxed{\frac{e\sqrt e-1}{e\sqrt e}} \]