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A system consisting of two components will continue to operate only as long as both components function. Suppose the joint pdf of the lifetimes (months) of the two components in a system is given by $f(x,y)=c[10-(x+y)] $ for $x>0$, $y>0$, $x+y<10$.

a. If the first component functions for exactly 3 months, what is the probability that the second functions for more than 2 months?

b. Suppose the system will continue to work only as long as both components function. Among 20 of these systems that operate independently of each other, what is the probability that at least half work for more than 3 months?

I am currently working on part a). I am trying to find the conditional probability by integrating $f(x,y)$ to find the marginal of $ X $ which I set as the first component. I was then thinking of dividing by $ f(x,y)$ to get the conditional probability function for $ Y $. Will this approach work?

For part b) I think I should use the binomial distribution and sum up 5 work+6 work+...+10 work. However, I still need the probability that a system works after 3 months and there might be an easier way. Any ideas?

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  • $\begingroup$ I don't know how to find the probability that a system lasts for over 3 months. Any ideas? $\endgroup$ – D. Wei Nov 30 '18 at 14:30
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First, the normalization constant is easily found to be $c = \left( \frac{ 10^3 }6\right)^{-1} = 0.006$, either by doing the integration of the density or by recognizing that the density describes the solid "corner" of a cube with side length of $10$.

(a) If the first component functions for exactly 3 months, what is the probability that the second functions for more than 2 months?

Denote the values of interest as $x_0 = 3$ and $y_0 = 2$. The requested probability is

$$\Pr\{ Y > y_0 ~|~ X = x_0 \} = \int_{y = y_0}^{10 - x_0} f_{Y|X=x_0}(y) \,\mathrm{d}y$$

where the conditional marginal density $f_{Y|X=x_0}$ is defined as

$$ f_{Y|X=x_0}(y) = \frac{ f(x_0, y) }{ \displaystyle \int_{y' = 0}^{10 - x_0} f(x_0, y') \,\mathrm{d}y' }$$ The $y'$ is just a dummy variable for integration.

This denominator is indeed $f_X(x_0)$ like you suggested, the marginal density of $X$ at $X = x_0$. The result is

$$\int_{y' = 0}^{10 - x_0} f(x_0, y') \,\mathrm{d}y' = c \int_{y' = 0}^{10 - x_0} (10 - x - y') \,\mathrm{d}y' = c \frac{ (10 - x_0)^2}2 = c \frac{49}2$$

Accordingly, the desired probability is \begin{align} \Pr\{ Y > y_0 ~|~ X = x_0 \} &= \int_{y = y_0}^{10 - x_0} f_{Y|X=x_0}(y) \,\mathrm{d}y \\ &= \frac1c \frac2{49} \int_{y = y_0}^{10 - x_0} f(x_0, y) \,\mathrm{d}y \\ &= \frac1c \frac2{49} \int_{y = y_0}^{10 - x_0} c \cdot (10 - x_0 - y) \,\mathrm{d}y \\ &= \frac2{49} \left( (10 - x_0)y - \frac{y^2}2 \right)\Big|_{y = y_0}^{y = 10-x_0} \\ &= \frac2{49} \left( \frac{(10 - x_0)^2}2 - (10 - x_0) y_0 + \frac{y_0^2}2 \right) \\ &= \frac2{49} \frac{25}2 \\ &= \frac{25}{49} \end{align}

(b) Among 20 of these systems that operate independently of each other, what is the probability that at least half work for more than 3 months?

The probability that a single system works for more than 3 months is

\begin{align} p &\equiv \Pr\{ X> 3 ~~\&~~ Y > 3\} \\ &= \int_{x = 3}^{10} \int_{y = 3}^{10 - x} c \cdot (10 - x - y) \,\mathrm{d}y \,\mathrm{d}x \\ &= c \int_{x = 3}^{10} \left( (10 - x)(10-x - 3) - \frac{(10-x)^2 - 3^2}2 \right) \,\mathrm{d}x \\ &= \ldots \text{some routine algebra} \ldots \\ &= c \frac{91}6 \\ & = \frac{91}{1000} \end{align} Note that any individual system has less than one tenth of a chance to last more than 3 months.

Finally, the probability that at least half of 20 work more than 3 months is the cumulative of a $\mathrm{Binomial}(20, p)$ with $p = 0.091$

$$\mathcal{P} \equiv \sum_{k = 10}^{20} {20 \choose k} p^k (1-p)^{20-k} = 1 - \sum_{k = 0}^9 {20 \choose k} p^k (1-p)^{20-k} \approx 3.04345 \cdot 10^{-6}$$

This sum has no general formula and must to be done via brute-force, either by consulting look-up table or Wolfram Alpha or the alike.

Not surprisingly, the requested probability $\mathcal{P}$, that at least half of 20 works more than 3 months, is vanishingly small because each individual system has not even a tenth of chance to "success".

Note that the $p = 0.091$ is kind of extreme so Normal approximation would not be good (unless $n$ is VERY large), and $n = 20$ is too small for Poisson approximation to work well, either.

Just for the record: Normal approximation with mean $\mu = n p$ and variance $\sigma^2 = np(1-p)$ with $n = 20$ and $p = \frac{91}{1000}$ is underestimating by two orders of magnitude. Meanwhile, Poisson approximation with parameter $\lambda = np$ overestimates by about 7 times.

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