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There is a system of pipes from one point to another. Pipe A is the start point, and connects left to right to Pipe B, C & D, which are in parallel and connect to the end point. The pipes can flow if and only if there is a path A to the end point. In essence, pipe A must work and at least one of B, C & D must work.

The likelihood of any pipe being blocked in $t$ amount of years is modelled by a $\text{Exp}(3)$ distribution, and all pipes are blocked independently of one another.

I have been asked to find the mean time for the system to not work, and if it is known that the system is working, what is the probability that B is not blocked?

First, I would like to find an expression for the likelihood of this system working at time $t$, here is my working:

A, B & C are the event that these pipes are not blocked

$$\begin{align} &\text{Let $Y$ be that the system works at a time $t$} \\ \therefore\quad\mathbb{P}(Y) &= \mathbb{P}(A \cap B) + \mathbb{P}(A \cap C) + \mathbb{P}(A \cap D) \\[1ex]&\text{Since all pipes block independently, and are identically distributed} \\ &= \mathbb{P}(A)\mathbb{P}(B) + \mathbb{P}(A)\mathbb{P}(C) + \mathbb{P}(A)\mathbb{P}(D) \\ &= 3\,\mathbb{P}(A)\mathbb{P}(B) \\[1ex] &{\text{Since A & B are modeled by the remainder of}\\\text{ an exponential distribution with $\lambda$ = 3}} \\ &= 3\cdot(1-(1-e^{-3t}))^{2} \\ &= 3\cdot(e^{-3t})^{2} \\ &= 3e^{-6t} \end{align}$$

For the mean time to failure, I did the following calculation:

\begin{align*} \mathbb{E}(Y) = \int_{0}^{\infty} te^{-6t} dt = \frac{1}{36} \end{align*}

Now, I am having difficulty understanding how to calculate $\mathbb{P}(B \vert Y)$

Here is my working so far, but I am almost certain it is wrong:

\begin{align*} \mathbb{P}(B \vert Y) &= \frac{\mathbb{P}(B \cap Y)}{\mathbb{P}(Y)} \\ \mathbb{P}(B \cap Y) &= \mathbb{P}(B \cap ((A \cap B) \cup (A \cap C) \cup (A \cap D)) \\ &= \mathbb{P}(A\cap B) + \mathbb{P}(B \cap A \cap C) + \mathbb{P}(B \cap A \cap D) \\ &= \mathbb{P}(A)\mathbb{P}(B) + \mathbb{P}(B)\mathbb{P}(A)\mathbb{P}(C) + \mathbb{P}(B)\mathbb{P}(A)\mathbb{P}(D) \\ &= (e^{-3t})^{2} + (e^{-3t})^{3} + (e^{-3t})^{3} \\ &= e^{-6t} + 2e^{-9t} \\ &= e^{-6t}(1 + 2e^{-3t}) \\ \therefore \mathbb{P}(B \vert Y) &= \frac{e^{-6t}(1 + 2e^{-3t})}{3e^{-6t}} = \frac{1 + 2e^{-3t}}{3} \end{align*}

Please let me know where I have messed up, thankyou so much for any help.

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3 Answers 3

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You need to apply inclusion exclusion to find $P(Y)$. You're double counting as it is. Luckily you can just use the fact that $P(Y) = P(A)P(B \cup C \cup D) = P(A)(1 - P(B^c \cap C^c \cap D^c ))$, where $B^c$ denotes the complement of $B$. To make computing $P(B | Y)$ easier notice that as $A$ must work, $P(B | Y) = P(B |B \cup C \cup D) = P(B)/P(B \cup C \cup D)$.

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Here is another approach to find the mean time that the system doesn't work, in case you're interested.

Let $T_a,T_b,T_c,T_d\sim \exp(3)$ denote the times that pipes $A,B,C,D$ become blocked and $T$ as the time that the system shuts down.

Notice $X=T_a\sim \exp(3)$ and $Y=\max\{T_b,T_c,T_d\}\sim f_Y$ where $$f_Y(y)=9e^{-3y}\left(1-e^{-3y}\right)^2\cdot 1_{[0,\infty)}(y)$$ Since $X,Y$ are independent $f_{X,Y}\equiv f_X \cdot f_Y$ and we have $$\begin{eqnarray*}\mathbb{E}(T) &=& \mathbb{E}(T|T_a>\max\{T_b,T_c,T_d\})\mathbb{P}(T_a>\max\{T_b,T_c,T_d\})+\mathbb{E}(T|T_a\leq\max\{T_b,T_c,T_d\})\mathbb{P}(T_a\leq\max\{T_b,T_c,T_d\}) \\ &=& \mathbb{E}(Y|X>Y)\mathbb{P}(X>Y)+\mathbb{E}(X|X\leq Y)\mathbb{P}(X\leq Y) \\ &=& \int_0^{\infty}\int_0^xyf_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x + \int_0^{\infty} \int_0^y xf_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y \\ &=& \frac{1}{4}\end{eqnarray*}$$

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The system fails by time $t$, if either pipe A fails, or all of pipes $B,C,D$ fail.

That is the event $\min\{T_A,\max\{T_B,T_C,T_D\}\}\leq t$.

So the cumulative distribution function for the time until system failure is is:

$$\begin{align}F_T(t) &= \mathsf P(\min\{T_A,\max\{T_B,T_C,T_D\}\}\leq t)\\&=1-\mathsf P(\min\{T_A,\max\{T_B,T_C,T_D\}\}>t)\\&=1-\mathsf P(T_A>t)\,\mathsf P(\max\{T_B,T_C,T_D\}>t)\\&=1-\mathsf P(T_A>t)\,(1-\mathsf P(\max\{T_B,T_C,T_D\}\leqslant t))\\&=1-\mathsf P(T_A> t)~(1-\mathsf P(T_B\leqslant t)\,\mathsf P(T_C\leqslant t)\,\mathsf P(T_D\leqslant t))\\&=(1-\mathrm e^{-3t}~(1-(1-\mathrm e^{-3t})^3))\,\mathbf 1_{0\leqslant t}\\&=(1-3\mathrm e^{-6t}+3\mathrm e^{-9t}+\mathrm e^{-12t})\mathbf 1_{0\leqslant t}\end{align}$$

The expected time until system failure may be found using that when random variable $T$ is most certainly non-negative, then we have: $\mathsf E(T)=\int_0^\infty 1-F_T(t)\,\mathrm d t$.

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