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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 ${位}_i,i=1,2$. A battery that is randomly chosen from the bin will be a type i battery with probability pi, $p_i,\underset{i=1}{\overset{2}{鈭�}}鈥�p_i=1$.

$X=$life time of a randomly chosen battery from the bin.

If a randomly chosen battery is still operating after $t$hours

So, the probability that it will still be operating after an additional $s$hours,

$P(X>s+t鈭�X>t)=\frac{P(X>s+t鈭�X>t)}{P(X>t)}$

$=\frac{P(X>s+t)}{P(X>t)}鈫�\left(\mathrm{I}\right)$

The selected battery may be of type 1 or of type2, so the numerator and the denominator of the equation (I) can be obtained using conditional probabilities as follows,

$P(X>s+t)=\left\{\begin{array}{l}P(X>s+t鈭�\text{Typel battery is selected})P\left(\text{Typel battery is selected}\right)+\\ P(X>s+t鈭�\text{Type2 battery is selected})P\left(\text{Type2 battery is selected}\right)\end{array}\right\}$

$=\left\{\begin{array}{l}P\left(X>s+t鈭�X~\text{Exponential}\left({位}_1\right)\right)P\left(\text{Typel battery is selected}\right)+\\ P\left(X>s+t鈭�X~\mathrm{Exponential}鈦�\left({位}_2\right)\right)P\left(\text{Type2 battery is selected}\right)\end{array}\right\}$

$=\left\{e^{鈭�{位}_1(s+t)}{p}_1+e^{鈭�{位}_2(s+1)}{p}_2\right\}鈫�\left(\text{II}\right)\left[\begin{array}{l}鈭�X~\mathrm{Exponential}鈦�\left(位\right)\\ 鈬�P(X>x)=e^{鈭�位x}\end{array}\right]$

So again,

$P(X>t)=\left\{\begin{array}{l}P(X>t鈭�\text{Typel battery is selected})P\left(\text{Typel battery is selected}\right)+\\ P(X>t鈭�\text{Type}2\text{battery is selected})P\left(\text{Type}2\text{battery is selected}\right)\end{array}\right\}$

$=\left\{\begin{array}{l}P\left(X>t鈭�X~\text{Exponential}\left({位}_1\right)\right)P\left(\text{Typel battery is selected}\right)+\\ P\left(X>t鈭�X~\text{Exponential}\left({位}_2\right)\right)P\left(\text{Type}2\text{battery is selected}\right)\end{array}\right\}$

$=\left\{e^{鈭�{位}_{1}t}{p}_1+e^{鈭�{位}_{2}t}{p}_2\right\}鈫�\left(\mathrm{III}\right)\left[\begin{array}{l}\text{Since}X~\mathrm{Exponential}鈦�\left(位\right)\\ 鈬�P(X>x)=e^{鈭�位x}\end{array}\right]$

Now substituting (II) & (III) in (I) we have the required probability,

$P(X>s+t鈭�X>t)=\frac{e^{鈭�{位}_1(s+t)}{p}_1+e^{鈭�{位}_2(s+t)}{p}_2}{e^{鈭�{位}_{1}t}{p}_1+e^{鈭�{位}_{2}t}{p}_2}$

The probability that it will still be operating after an additionals hours will be$P(X>s+t鈭�X>t)=\frac{e^{鈭�{位}_1(s+t)}{p}_1+e^{鈭�{位}_2(s+t)}{p}_2}{e^{鈭�{位}_{1}t}{p}_1+e^{鈭�{位}_{2}t}{p}_2}$