Discussion
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
\[ f(x, y) = \begin{cases} \frac{1}{10} e^{-\frac{1}{2}(y + 3 - x)}, & 5 \lt x \lt 10,\ y \gt x - 3, \\ 0, & \text{otherwise.} \end{cases} \]
The probability, that component B fails before component A, is equal to
Answer: A
The probability is \(\frac{e\sqrt{e} - 1}{e\sqrt{e}}\), that is \(1 - e^{-3/2}\).
For fixed \(x\) in \((5, 10)\), the density in \(y\) is \(\frac{1}{10}e^{-\frac{1}{2}(y - (x - 3))}\) for \(y \gt x - 3\). Integrating over \(y\) gives \(\frac{1}{5}\), so \(X \sim U(5, 10)\), and given \(X = x\), \(Y - (x - 3)\) is exponential with mean 2.
Component B fails first when \(Y \lt X\), that is \(Y - (x - 3) \lt 3\):
\[ \begin{aligned} P(Y \lt X \mid X = x) &= 1 - e^{-\frac{3}{2}} \\ P(Y \lt X) &= \int_{5}^{10} \frac{1}{5}\left(1 - e^{-\frac{3}{2}}\right)dx = 1 - e^{-\frac{3}{2}} \end{aligned} \]
Since \(e^{3/2} = e\sqrt{e}\), this is \(\frac{e\sqrt{e} - 1}{e\sqrt{e}}\), about 0.78. \(\frac{1}{e\sqrt{e}}\) is the complementary probability that A fails first, and the two options with \(e\sqrt{e} + 1\) in the denominator have no basis in the integral.
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