# Inconsistent answers from inferring probability of success from probability of failure

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.

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.
• 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 $0.02\cdot 0.997$ – Ross Millikan May 22 at 3:14
• $${\hspace{4.25ex}\mathsf P(A\cap B)~+~\mathsf P(A\cap B^\complement)~+~\mathsf P(A^\complement\cap B)~+~\mathsf P(A^\complement\cap B^\complement)\\=(0.02\cdot 0.03)+(0.02\cdot0.97)+(0.98\cdot0.03)+(0.98\cdot 097)\\=1}$$ – Graham Kemp May 22 at 4:45