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# There are two types of batteries in a bin. When in use, type i batteries last (in hours) an exponentially distributed time with rate$\lambda _ { i } , i = 1,2$. A battery that is randomly chosen from the bin will be a type i battery with probability$p _ { i } , \sum _ { i = 1 } ^ { 2 } p _ { i } = 1$If a randomly chosen battery is still operating after t hours of use, what is the probability that it will still be operating after an additional s hours?

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Define $X$ as the random variable that marks the time in which randomly chosen battery operates. Define events $A$ that the first battery has been chosen and $B$ that the second battery has been chosen. We are required to calculate $P(X > t+s | X >t)$. Using the law of the total probability, we have that

\begin{align} P(X > t+s | X >t) &= P(X > t+s| X > t, A)P(A|X>t) \\ &+ P(X > t+s| X > t, B)P(B|X>t) \end{align}

Now, if we are given that the first battery has been chosen, we have that $X \sim \text{Expo}(\lambda_1)$, so we can use memoryless property of exponential distribution to obtain that

\begin{align*} P(X > t+s| X > t, A) = P(X>s|A) = e^{-\lambda_1s} \end{align*}

On the other hand, use Bayesian formula to obtain that

$\begin{gather*} P(A|X>t) = \frac{P(X>t|A)P(A)}{P(X>t)} = \frac{P(X>t|A)P(A)}{P(X>t|A)P(A) + P(X>t|B)P(B)} \\ = \frac{e^{-\lambda_1t}p_1}{e^{-\lambda_1t}p_1 + e^{-\lambda_2t}p_2} \end{gather*}$

Similarly we have that

\begin{align*} P(X > t+s| X > t, B) = P(X>s|B) = e^{-\lambda_2s} \end{align*}

and

$\begin{gather*} P(B|X>t) = \frac{P(X>t|A)P(B)}{P(X>t)} = \frac{P(X>t|B)P(B)}{P(X>t|A)P(A) + P(X>t|B)P(B)} \\ = \frac{e^{-\lambda_2t}p_2}{e^{-\lambda_1t}p_1 + e^{-\lambda_2t}p_2} \end{gather*}$

Finally, the required probability (from the expression (1)) is

\begin{align*} e^{-\lambda_1s} \cdot \frac{e^{-\lambda_1t}p_1}{e^{-\lambda_1t}p_1 + e^{-\lambda_2t}p_2} + e^{-\lambda_2s} \cdot \frac{e^{-\lambda_2t}p_2}{e^{-\lambda_1t}p_1 + e^{-\lambda_2t}p_2} \end{align*}

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