A system contains 8 chips. The lifetime of each chip has a Weibull probability law $\left(P\left[(t,\infty)\right]=\mathrm{e}^{-(\lambda t)^k}\;\style{font-family:inherit;}{\text{ for }}\;0\leq t\right)$ w/ $k = 2$. Find the probability that at least 2 chips are functioning after $2/\lambda$ seconds.
I found a couple of answers to this q (namely, 0.008727 & 0.0095) but let me explain where I think they made their mistake:
From what I've read about Weibull, for 2 parameters, it gives us the probability that a device will MALFUNCTION after a certain time has passed; I suspect they assumed otherwise.
My solution follows.
Let...
- $q=P\left[\style{font-family:inherit;}{\text{"Chip malfunctions after }}2/\lambda \style{font-family:inherit;}{\text{ seconds"}}\right]$.
- $p=P\left[\style{font-family:inherit;}{\text{"Chip functions after }}2/\lambda \style{font-family:inherit;}{\text{ seconds"}}\right]=1-q$.
- random var $X=$ "# functioning chips after $2/\lambda$ seconds".
Therefore...
- $q=P\left[(2/\lambda,\infty)\right]=\mathrm{e}^{-\left(\lambda \left(\frac{2}{\lambda}\right)\right)^{(2)}}=\mathrm{e}^{-4}$
- $p=1-\mathrm{e}^{-4}$
- $P\left[X\geq2\right] = 1 - P\left[X < 2\right] = 1 - \left(P\left[X = 0\right] + P\left[X = 1\right]\right)\\ = 1 - \left(\bigl(\begin{smallmatrix} 8\\0 \end{smallmatrix}\bigr)q^{8-0}p^0\right) - \left(\bigl(\begin{smallmatrix} 8\\1 \end{smallmatrix}\bigr)q^{8-1}p^1\right) = 1 - q^8 - 8q^7(1-q) = 1 - 8q^7 + 7q^8\\ \Longrightarrow P\left[X\geq2\right] = 0.99999999999455(7)$
So, were they mistaken, or am I?
Let $T$ denote the lifetime of a typical chip. The event $\{T>2/\lambda\}$ means that lifetime of a chip is greater that $2/\lambda$ and it is functioning at least $2/\lambda$ seconds. The event "Chip malfunctions after 2/ seconds" means that any time after $2/\lambda$ this chip is malfunctioning. This is exactly the event $\{T\leq 2/\lambda\}$.
So
$p=\mathbb P\left[\style{font-family:inherit;}{\text{"Chip functions after }}2/\lambda \style{font-family:inherit;}{\text{ seconds"}}\right]=\mathbb P\left[T>2/\lambda\right]=e^{-4}$.
Right answer is $0{,}008727978$.