A machine has $5$ components and needs at least 3 working components to function. Suppose that their lifetimes are independent exponential(1). Find the density function for the time to failure $T$.


Let $T$ be the time to failure. We need $f_T(t)$. Lets find $P(T \leq t)$. Let $X_i$ be lifetime of each of five components and each $X_i$ is exp(1). Thus, $f_{X_i} (x) = e^{-x} $ for $x>0$.

Notice that P(one component not working) is $P(X_i > t ) = e^{-t} $

So, here is my confusion. Isnt $T$ a binomial random variable with $n=5$ and $p=e^{-t}$? and so

$$ P(number \; of \; components \; not \; working < 3) = (1-e^{-t})^5 + 5e^{-t}(1-e^{-t})^4 + {5 \choose 2} e^{-2t} (1-e^{-t})^3 $$

Can we consider this to be our distribution and so $f_T(t)$ would just be the derivative of the above expression?

Best regards,



No, $T$ cannot be binomial because it is a time to failure (of the system). If you want to count the number of components still operating at a given time, that at least is compatible with a discrete distribution with finite support.

If each component's lifetime is independent and identically exponentially distributed with mean $1$, then the time to failure of the system occurs when the third component fails (the system can operate with two failures, but not three). So this should immediately suggest that we consider the order statistics. In other words, let $X_{(1)}, X_{(2)}, X_{(3)}, X_{(4)}, X_{(5)}$ represent the failure times of the five components in ascending order, so in particular $X_{(i)} \le X_{(j)}$ for any $i < j$. Then $T = X_{(3)}$, the time to the third failure.

So how do we compute $\Pr[T \le t] = \Pr[X_{(3)} \le t]$? What this means is that at least three failures (but possibly more) have occurred before time $t$. If only two failures have happened by time $t$, then the event $X_{(3)} \le t$ did not occur. So we note by the independence of individual failure times $$\Pr[X_{(3)} \le t] = \binom{5}{3} \Pr[X \le t]^3 \Pr[X > t]^2 + \binom{5}{4} \Pr[X \le t]^4 \Pr[X > t] + \binom{5}{5} \Pr[X \le t]^5.$$ Why? Partition the LHS probability into three cases:

  1. Exactly 3 components have failed by time $t$ and 2 have not;
  2. Exactly 4 components have failed by time $t$ and 1 has not;
  3. All 5 components have failed by time $t$.

Since each component's lifetime is independent, the first case gives us $\Pr[X \le t]^3 \Pr[X > t]$ for a given ordering of components. But there are $\binom{5}{3}$ ways to choose the three components that fail. Similarly, we must account for the $\binom{5}{4}$ ways to choose four components that fail.

Then, once you have computed the above, we simply take the derivative to find the density of $T$. Your computation agrees with mine, except for some minor errors and the confusion between a time-to-event variable and a counting variable, as I pointed out at the start.


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