= Solution
{c}
\b[Conditioning on the observed time.] The native PDF defines $\tau_2$ as the first production time and then conditions on that production occurring at $\tau_2$. If the intended information is its numerical value $\tau_2=t$, the answer depends on $t$. We first compute this <conditional probability> using a <conditional probability density>; conditioning on an exact value of a continuous <random variable> means a density ratio, since $\mathbb P(\tau_2=t)=0$.
Before $\tau_2$, no molecules of the second <chemical species> are present, so the joint-removal <chemical reaction> cannot occur. Molecules of the first <chemical species> arrive as a rate-$2$ <Poisson process>, and each independently starts a clock with a rate-$1/2$ <exponential distribution> for producing the first molecule of the second <chemical species>. Define the unnormalized <probability mass function>
$$
q_n(t)=\mathbb P\bigl(A_1(t)=n,\ \tau_2>t\bigr),\qquad
F(z,t)=\sum_{n=0}^\infty q_n(t)z^n.
$$
The <chemical master equation> before the first production, with production treated as killing, is
$$
q_n'=2q_{n-1}-\left(2+\frac n2\right)q_n,
\qquad q_n(0)=\mathbf1_{\{n=0\}}.
$$
Multiplying by $z^n$ and summing gives a <partial differential equation> for the <probability generating function>:
$$
F_t=2(z-1)F-\frac z2F_z,\qquad F(z,0)=1.
$$
Its solution, verifiable by direct <differentiation>, is
$$
F(z,t)=\exp\bigl[-2t+m(t)z\bigr],
\qquad m(t)=4(1-e^{-t/2}).
$$
Thus $q_n(t)=e^{-2t}m(t)^n/n!$, the <survival function> is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the <conditional distribution> of $A_1(t)$ given survival to $t$ is a <Poisson distribution> of <mean> $m(t)$.
Production in $(t,t+dt)$ from <Markov chain state> $n$ has joint <probability> $(n/2)q_n(t)\,dt+o(dt)$. Therefore the <probability density function> of $\tau_2$ is
$$
f_{\tau_2}(t)=\frac12\sum_{n\geq1}nq_n(t)
=\frac12m(t)S(t),\qquad t>0.
$$
The first production leaves $A_1$ unchanged. Taking the density ratio gives a <size-biased distribution> of the surviving count:
$$
\mathbb P\bigl(A_1(\tau_2)=n\mid\tau_2=t\bigr)
=\frac{nq_n(t)}{m(t)S(t)}
=e^{-m(t)}\frac{m(t)^{n-1}}{(n-1)!},\qquad n\geq1.
$$
In particular, under this reading the requested <conditional probability> is
$$
\boxed{\mathbb P\bigl(A_1(\tau_2)=2\mid\tau_2=t\bigr)
=4(1-e^{-t/2})\exp\!\bigl[-4(1-e^{-t/2})\bigr],\quad t>0.}
$$
The shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production <reaction propensity function> is larger.
\b[Without observing the numerical time.] If the conditioning merely repeats the definition of $\tau_2$, the requested <probability> is instead the unconditional count at that <stopping time>. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in <Markov chain state> $n$. The <competing exponential clocks> imply
$$
\mathbb P\bigl(A_1(\tau_2)=2\bigr)
=\frac{2}{2+1/2}\frac{1}{2+1}
=\boxed{\frac4{15}}.
$$
The first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with <probability> one, as is also clear from $S(t)\to0$. As a consistency check, averaging the time-conditioned answer against $f_{\tau_2}$ gives
$$
\int_0^\infty m(t)e^{-m(t)}\frac{m(t)}2e^{-2t+m(t)}\,dt
=8\int_0^\infty e^{-2t}(1-e^{-t/2})^2\,dt
=\frac4{15}.
$$
These are two distinct, explicitly specified interpretations of the printed conditioning.
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