In the stationary regime, $A+B$ is distributed like a size-biased version $S$ of the lifetime of a component (more about this later) and $(A,B)$ is distributed like $(US,(1-U)S)$, where $U$ is uniform on $[0,1]$ and independent of $S$. Thus, to compute $p=\mathbb P(B\geqslant b\mid A=a)$, one needs to know the density $f_{A,B}$ of the joint distribution of $(A,B)$, since then $p=q(b)/q(0)$, where $ q(z)=\int_{y\geqslant z}f_{A,B}(a,y)\mathrm dy. $ The distribution of $(A,B)$ is entirely determined by the above if one adds a description of the distribution of $S$: this is the unique distribution such that $\mathbb E(\varphi(S))=\mathbb E(T\varphi(T))/\mathbb E(T)$ for every bounded measurable function $\varphi$.
Thus, to compute $f_{A,B}$, consider any bounded measurable function $\varphi$ and use the identity $ (*)=\iint\varphi(x,y)f_{A,B}(x,y)\mathrm dx\mathrm dy=\mathbb E(\varphi(A,B))=\mathbb E(\varphi(US,(1-U)S)), $ which yields $ (*)=\mathbb E(T\varphi(UT,(1-U)T))/\mathbb E(T). $ Measuring $T$ in hundreds of hours, $T$ is uniform on $[1,2]$ hence $\mathbb E(T)=\frac32$ and $ (*)=\tfrac23\int_1^2t\int_0^1\varphi(ut,(1-u)t)\mathrm du\mathrm dt. $ The change of variables $(x,y)=(ut,(1-u)t)$ with Jacobian $\mathrm dx\mathrm dy=t\mathrm dt\mathrm du$ yields $ (*)=\tfrac23\iint\varphi(x,y)\mathbf 1_{1\leqslant x+y\leqslant2}\mathbf 1_{x\geqslant0,y\geqslant0}\mathrm dx\mathrm dy, $ which implies that $ f_{A,B}(x,y)=\tfrac23\mathbf 1_{1\leqslant x+y\leqslant2}\mathbf 1_{x\geqslant0,y\geqslant0}. $ In other words, $(A,B)$ is uniform on the inside of the quadrilateral with vertices $(1,0)$, $(2,0)$, $(0,2)$ and $(0,1)$.
In particular, conditionally on $[A=a]$ for some $a$ in $[0,1]$, $B$ is uniformly distributed on $[1-a,2-a]$ and, for every $b$ in $[1-a,2-a]$, $p=\mathbb P(B\geqslant b\mid A=a)=2-a-b$. Your case is $a=\frac9{10}$, $b=\frac12$, hence $p=\frac35$.
Note finally that $A$ and $B$ are identically distributed and not independent (this is always the case) and that, in the present case, their common distribution has density $\tfrac23(\mathbb 1_{0\leqslant x\leqslant1}+(2-x)\mathbb 1_{1\lt x\leqslant2}), $ hence $\mathbb E(A)=\mathbb E(B)=\frac79$. In the general case, $ f_{A,B}(x,y)=\frac{f_T(x+y)}{\mathbb E(T)}\mathbf 1_{x\geqslant0,y\geqslant0}, $ where $f_T$ is the density of $T$. Furthermore, $\mathbb E(A)=\mathbb E(B)$ and their common value is $\frac12\mathbb E(S)$ where $\mathbb E(S)=\mathbb E(T^2)/\mathbb E(T)$, hence $\mathbb E(S)\gt\mathbb E(T)$ as soon as the distribution of $T$ is not degenerate.