A capacitor C_1 of capacitance 1 and a capacitor C_2 of capacitance...

Let the common battery voltage be $V_0$. The initial voltage across both capacitors is $V_0$.

During discharge, the current at time $t$ is given by $I(t) = \dfrac{V_0}{R} e^{-t/RC}$.

At $t = 0$, the initial current in both circuits is $I(0) = \dfrac{V_0}{R}$. Since $V_0$ and $R$ are the same for both circuits, the initial currents are equal and non-zero.

The charge on a discharging capacitor at time $t$ is $q(t) = q_0 e^{-t/RC}$.

The time $t$ taken to lose $50\%$ of the initial charge is obtained by setting $q(t) = 0.5 q_0$, which gives $t = RC \ln 2$.

Since $C_1 < C_2$, the time constant for the first circuit is smaller ($R C_1 < R C_2$).

Thus, $t_1 < t_2$, meaning $C_1$ loses $50\%$ of its initial charge sooner than $C_2$.