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The probability in question is $1-p$, where $p$ is the probability that we will hit the negative $x$-semiaxis before hitting the positive $x$-semiaxis. Next, $p$ is the probability that (we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis, and then we will hit the negative $x$-semiaxis before hitting the positive $x$-semiaxis). So, by the strong Markov property and the symmetry, \begin{equation*} p=q\,\tfrac12, \end{equation*} where $q$ is the probability that we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis. In turn, \begin{equation} q=P(\tau_a<\tau_b)=\int_0^\infty P(\tau_a<t)\,dP(\tau_b<t), \tag{1} \end{equation} where $\tau_a$ and $\tau_b$ are independent random variables (r.v.'s) such that for any real $c$ the distribution of the r.v. $\tau_c$ is that of the time for a standard Brownian motion (starting at $0$) to first reach point $c$: \begin{equation*} P(\tau_c<t)=2(1-\Phi(c/\sqrt t)) \end{equation*} for $t>0$, by the reflection principle, where $\Phi$ is the standard normal distribution function. Thus, by (1), the probability in question is \begin{equation} 1-q/2= 1-\int_0^\infty (1-\Phi(a/\sqrt t))\varphi(b/\sqrt t)\frac{b\,dt}{t^{3/2}}, \end{equation} where $\varphi:=\Phi'$. It may or may not be possible to evaluate the latter integral in closed form.