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Suppose we bought a hard drive. With probability $p$ we've bought a good product, and with probability $1 - p$ it is defective.

  • If the hard drive is good, the time to failure is exponentially distributed with parameter $\lambda_1$.
  • If the hard drive is defective, the time to failure is exponentially distributed with parameter $\lambda_2$.

Now given that our hard drive has been functioning for time $T$, what is the probability that it is defective?

I tried to compute $$P(defective | t \ge T) = \frac{P(defective \cap t \ge T)}{P(t \ge T)},$$ but I cannot think of a way to compute the numerator. How can I compute the probability of the intersection?

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    The numerator is slightly simpler than the denominator but the calculations are similar. How would you calculate the denominator? – 2017-02-09
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    @Henry The denominator would be $pe^{-\lambda_1 T} + (1-p)e^{-\lambda_2 T}$. – 2017-02-09
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    That denominator is $P(\text{good} \cap t \ge T) + P(\text{defective} \cap t \ge T)$. So the numerator for "defective" is the second part of that: $(1-p)e^{-\lambda_2 T}$ – 2017-02-09
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    @Henry Ah, I guess for some reason I couldn't think. Thanks! – 2017-02-09

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As a summary of the comments:

$P(\text{defective} \mid t \ge T)=\dfrac{P(\text{defective} \cap t \ge T)}{P(\text{good} \cap t \ge T) + P(\text{defective} \cap t \ge T)} = \dfrac{(1-p)e^{-\lambda_2 T}}{pe^{-\lambda_1 T} + (1-p)e^{-\lambda_2 T}}$