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Question

There are two types of batteries in a bin. When in use, type i batteries last (in hours) an exponentially distributed time with rate

λi,i=1,2\lambda _ { i } , i = 1,2

. A battery that is randomly chosen from the bin will be a type i battery with probability

pi,i=12pi=1p _ { 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?

Solution

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

P(X>t+sX>t)=P(X>t+sX>t,A)P(AX>t)+P(X>t+sX>t,B)P(BX>t)\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 XExpo(λ1)X \sim \text{Expo}(\lambda_1), so we can use memoryless property of exponential distribution to obtain that

P(X>t+sX>t,A)=P(X>sA)=eλ1s\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

P(AX>t)=P(X>tA)P(A)P(X>t)=P(X>tA)P(A)P(X>tA)P(A)+P(X>tB)P(B)=eλ1tp1eλ1tp1+eλ2tp2\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

P(X>t+sX>t,B)=P(X>sB)=eλ2s\begin{align*} P(X > t+s| X > t, B) = P(X>s|B) = e^{-\lambda_2s} \end{align*}

and

P(BX>t)=P(X>tA)P(B)P(X>t)=P(X>tB)P(B)P(X>tA)P(A)+P(X>tB)P(B)=eλ2tp2eλ1tp1+eλ2tp2\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

eλ1seλ1tp1eλ1tp1+eλ2tp2+eλ2seλ2tp2eλ1tp1+eλ2tp2\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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