The probability that the k -th block of a computer fails…
The probability that the $k$-th block of a computer fails during a time $T$ is $p_k$ $(k=1,2,\dots,n)$. Determine the probability that during this time at least one block fails, if the blocks work independently.
Explanation of Method
"At least one fails" is the opposite of "no block fails", i.e. "all blocks work". Since the blocks are independent, the probability that all of them work is the product of the probabilities $1-p_k$. Subtracting from $1$ gives the answer.
Step 1: Events
$A_k$ = "block $k$ fails", $P(A_k)=p_k$, independent. We want $P(A_1+A_2+\dots+A_n)$.
Step 2: The opposite event
By De Morgan, $\overline{A_1+\dots+A_n}=\overline{A_1}\cdots\overline{A_n}$ ("all blocks work"). By independence:
\[ P(\overline{A_1}\cdots\overline{A_n}) = (1-p_1)(1-p_2)\cdots(1-p_n). \]
Step 3: The answer
\[ \boxed{P(\text{at least one fails}) = 1-\prod_{k=1}^{n}(1-p_k).} \]
Step 4: Equal probabilities and small $p$
- If all $p_k=p$: $P=1-(1-p)^n$.
- If all $p_k$ are small: $P\approx p_1+p_2+\dots+p_n$ (the products of two or more small numbers are negligible).
Example: $n=100$ blocks, each with $p=0.01$: $P=1-0.99^{100}\approx0.634$. Although each block is very reliable, the whole machine fails with probability almost $\frac23$.
Result
| Answer | |
|---|---|
| Probability | $1-(1-p_1)(1-p_2)\cdots(1-p_n)$ |
| Equal $p_k=p$ | $1-(1-p)^n$ |