Suppose I have a very large population of parts with a known 20% failure rate. Suppose I randomly select 5 samples and inspect for the known failure.

I calculate that the probability of all five samples passing is =0.8*0.8*0.8*0.8*0.8=0.328 P( 0.8 ∩ 0.8 ∩ 0.8 ∩ 0.8 ∩ 0.8 ) = 0.328 I believe this is correct.

If I reverse the thought process and ask what is the probability of one sample failing, I do: =0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1.0 P( 0.2 ∪ 0.2 ∪ 0.2 ∪ 0.2 ∪ 0.2 ) = 1.0 I know this is wrong.

I know using the OR function in this matter is a fallacy, because I know that I would not be guaranteed to find a failure. What I don't know is WHY? Why is the OR function incorrect in this case?

Please let me know if you see any other errors in my math.

Note: this is not for homework. I finished college statistics 12 years ago.


The primary issue is that $P(A \cup B) = P(A) + P(B)$ only if $A$ and $B$ are mutually exclusive. More generally, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ with a more general form for finding $P(\bigcup_{i=1}^k A_i)$. This is known as the inclusion-exclusion rule.

To reverse the process, you can realize that the probability of at least one sample failing is the complement of none of the samples failing (this is one version of De Morgan's Law). Let $A_i$ be the event that the $i^{th}$ sample fails. \begin{align*} P(A_1 \cup A_2 \cup \cdots A_5) &= 1 - P((A_1 \cup A_2 \cup \cdots \cup A_5)^c) \\ &= 1 - P(A_1^c \cap A_2^c \cap \cdots \cap A_5^c) \\ &= 1 - P(A_1^c)P(A_2^c)\cdots P(A_5^c) \\ &= 1 - 0.8^5 = .67232 \end{align*}

As a final comment, you could look into the Binomial Distribution for a more flexible approach. Suppose you select $n$ samples, and the pass rate is $p = 0.8$. If $X$ is the number of samples which pass inspection, then $X \sim \text{Binomial}(n, 0.8)$ with probability mass function $$P(X=x) = \binom{n}{x}\ 0.8^x\ 0.2^{n-x}$$ This general formula allows you to easily answer both questions posed above, as well as any other probability questions you may have.


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