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: Understand the required probability.
Component B fails before component A means
\[ Y<X \] So, we need to find
\[ P(Y<X) \]

Step 2: Identify the region of integration.
The joint density is non-zero when
\[ 5<x<10 \] and
\[ y>x-3 \] Also, for component B to fail before component A,
\[ y<x \] Therefore, for each \(x\), the limits of \(y\) are
\[ x-3<y<x \] and the limits of \(x\) are
\[ 5<x<10 \]

Step 3: Set up the double integral.
Thus,
\[ P(Y<X) = \int_{5}^{10}\int_{x-3}^{x} \dfrac{1}{10}e^{-\frac12(y+3-x)}\,dy\,dx \]

Step 4: Evaluate the inner integral.
Let
\[ u=y+3-x \] Then,
\[ du=dy \] When
\[ y=x-3, \] we get
\[ u=0 \] When
\[ y=x, \] we get
\[ u=3 \] Therefore, the inner integral becomes
\[ \int_{x-3}^{x} \dfrac{1}{10}e^{-\frac12(y+3-x)}\,dy = \dfrac{1}{10}\int_{0}^{3}e^{-\frac{u}{2}}\,du \] Now,
\[ \int e^{-\frac{u}{2}}du=-2e^{-\frac{u}{2}} \] So,
\[ \dfrac{1}{10}\int_{0}^{3}e^{-\frac{u}{2}}\,du = \dfrac{1}{10}\left[-2e^{-\frac{u}{2}}\right]_{0}^{3} \] \[ = \dfrac{1}{10}\left(-2e^{-\frac32}+2\right) \] \[ = \dfrac{1}{5}\left(1-e^{-\frac32}\right) \]

Step 5: Evaluate the outer integral.
Now,
\[ P(Y<X) = \int_{5}^{10} \dfrac{1}{5}\left(1-e^{-\frac32}\right)\,dx \] Since the integrand is constant,
\[ P(Y<X) = \dfrac{1}{5}\left(1-e^{-\frac32}\right)(10-5) \] \[ = 1-e^{-\frac32} \] Now,
\[ e^{\frac32}=e\sqrt{e} \] Therefore,
\[ 1-e^{-\frac32} = 1-\frac{1}{e\sqrt{e}} \] \[ = \frac{e\sqrt{e}-1}{e\sqrt{e}} \]

Step 6: Final conclusion.
Hence, the required probability is
\[ \boxed{\frac{e\sqrt{e}-1}{e\sqrt{e}}} \] Therefore, the correct option is
\[ \boxed{(A)} \]