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Alright so I was working on my previous post and stumbled into a problem. Say the $P(A$) failing is $0.02$, which translates to $2\%$ failure rate. Say the P(B) failing is 0.003, which translates to $0.3\%$ failure rate. Assuming the event of $A\cap B$ is mutually inclusive and independent, then $P=A\cdot B$ in terms of probability of failure. $$P(A\cap B)=P(A)\cdot P(B) \\ 0.02\cdot 0.003 \implies 6\times10^{-4}\implies P=0.00006 \ \text{or} \ 0.006\% \ \text{chance of failure }$$

Now, in terms of success I should be able to say $P(A)=2\%$ implies a success rate of $98\%$. Likewise, $P(B)=0.3\%$ implies a success rate of $99.7\%$. When I do the same exact operations in terms of success now, I get $$P(A\cap B)=P(A)\cdot P(B) \\ 0.98\cdot 0.997\implies P=0.97706 \ \text{or} \ 97.706\% \ \text{chance of success}$$

It doesn't matter what A and B represent, my issue is why the numbers don't match up. I should be able to say that, if $x$ = 100% and $y$=probability of failure, then $x-y$ = probability of success. A $2\%$ failure right literally implies a $98\%$ success rate and a $0.3\%$ failure rate literally implies a $97.706\%$ success rate! But thats not happening here.

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You compute the probability of both $A$ and $B$ failing versus the probability of both $A$ and $B$ succeeding. You ignore the cases when one of $A,B$ fails and the other succeeds.

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  • $\begingroup$ Ah, so how do I account for when one or the other fails or both fails? $\endgroup$ – Lex_i May 22 at 2:43
  • $\begingroup$ You have to decide whether A failing and B succeeding is success or failure. In either case, the probability of this is figured the same way, as $0.02\cdot 0.997$ $\endgroup$ – Ross Millikan May 22 at 3:14
  • $\begingroup$ $${\hspace{4.25ex}\mathsf P(A\cap B)~+~\mathsf P(A\cap B^\complement)~+~\mathsf P(A^\complement\cap B)~+~\mathsf P(A^\complement\cap B^\complement)\\=(0.02\cdot 0.03)+(0.02\cdot0.97)+(0.98\cdot0.03)+(0.98\cdot 097)\\=1}$$ $\endgroup$ – Graham Kemp May 22 at 4:45

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