Q.5.18 - A First Course in Probability

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 ${\lambda }_i,i=1,2$. A battery that is randomly chosen from the bin will be a type i battery with probability pi, $p_i,\sum _{i=1}^2 p_i=1$. 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? 

Step-by-Step Solution

Answer

The probability that it will still be operating after an additional s hours will be $P(X>s+t\mid X>t)=\frac{e^{-{\lambda }_1(s+t)}{p}_1+e^{-{\lambda }_2(s+t)}{p}_2}{e^{-{\lambda }_{1}t}{p}_1+e^{-{\lambda }_{2}t}{p}_2}$.

Step 1: Given information

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 pi, $p_i,\sum _{i=1}^2 p_i=1$.

Step 2: Solution

$X=$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,

$P(X>s+t\mid X>t)=\frac{P(X>s+t\cap X>t)}{P(X>t)}$

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

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,

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

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

$=\left\{e^{-{\lambda }_1(s+t)}{p}_1+e^{-{\lambda }_2(s+1)}{p}_2\right\}\to \left(\text{ II }\right)\left[\begin{array}{l}\because X~\mathrm{Exponential}\left(\lambda \right)\\ \Rightarrow P(X>x)=e^{-\lambda x}\end{array}\right]$

Step 3: Solution

So again,

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

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

$=\left\{e^{-{\lambda }_{1}t}{p}_1+e^{-{\lambda }_{2}t}{p}_2\right\}\to \left(\mathrm{III}\right)\left[\begin{array}{l}\text{ Since }X~\mathrm{Exponential}\left(\lambda \right)\\ \Rightarrow P(X>x)=e^{-\lambda x}\end{array}\right]$

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

$P(X>s+t\mid X>t)=\frac{e^{-{\lambda }_1(s+t)}{p}_1+e^{-{\lambda }_2(s+t)}{p}_2}{e^{-{\lambda }_{1}t}{p}_1+e^{-{\lambda }_{2}t}{p}_2}$

Step 4: Final answer

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