A device working during a time t consists of three parts…
A device working during a time $t$ consists of three parts, each of which may fail during this time independently of the others. If any part fails, the whole device stops working. The probabilities that the parts do not fail during time $t$ are $p_1=0.8$, $p_2=0.9$, $p_3=0.7$. Find the reliability of the whole device.
Explanation of Method
The reliability of the device is the probability that it works without failure during time $t$. The device works iff all three parts work (a "series" system), and the parts are independent, so by the multiplication rule the reliability is the product $p_1p_2p_3$.
Step 1: Events
$A_k$ = "part $k$ does not fail during time $t$", $P(A_k)=p_k$. The device works iff $A_1A_2A_3$ occurs.
Step 2: Multiplication rule for independent events
\[ \boxed{P = p_1p_2p_3 = 0.8\cdot0.9\cdot0.7 = 0.504.} \]
Step 3: Interpretation
Although each part is fairly reliable, the device works only about half the time: $P(\text{failure}) = 1-0.504=0.496$. The reliability of a series system is always smaller than that of its weakest part ($0.7$).
Step 4: Which part matters most?
| If this part is made perfect | New reliability |
|---|---|
| part 1 ($0.8\to1$) | $0.9\cdot0.7=0.63$ |
| part 2 ($0.9\to1$) | $0.8\cdot0.7=0.56$ |
| part 3 ($0.7\to1$) | $0.8\cdot0.9=0.72$ |
Improving the weakest part gives the biggest gain.
Result
| Answer | |
|---|---|
| Reliability | $0.8\cdot0.9\cdot0.7=0.504$ |