Discussion
Suppose the lifetimes of systems A and B are independent and follow exponential distributions with failure rates 3 and 2 respectively. Then the probability that system A fails before system B is :
Answer: C
The probability that A fails first is 0.6.
For independent exponential lifetimes with rates \(\lambda_{A}\) and \(\lambda_{B}\),
\[ \begin{aligned} P(T_{A} \lt T_{B}) &= \int_{0}^{\infty} \lambda_{A} e^{-\lambda_{A}t} e^{-\lambda_{B}t}\, dt \\ &= \frac{\lambda_{A}}{\lambda_{A} + \lambda_{B}} = \frac{3}{3 + 2} = 0.6 \end{aligned} \]
The system with the higher failure rate is more likely to fail first, so the answer must exceed 1/2. The value 0.4 is \(\lambda_{B}/(\lambda_{A} + \lambda_{B})\), the probability that B fails first. 0.5 would need equal rates, and 1 would need B never to fail.
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