  Which state does lose less quantum information between GHZ and W states when
they are prepared for two-party quantum teleportation through noisy channel? We
address this issue by solving analytically a master equation in the Lindbald
form with introducing the noisy channels which makes the quantum channels to be
mixed states. It is found that the answer of the question is dependent on the
type of the noisy channel. If, for example, the noisy channel is ($L_{2,x}$,
$L_{3,x}$, $L_{4,x}$)-type where $L's$ denote the Lindbald operators, GHZ state
is always more robust than W state, i.e. GHZ state preserves more quantum
information. In, however, ($L_{2,y}$, $L_{3,y}$, $L_{4,y}$)-type channel the
situation becomes completely reversed. In ($L_{2,z}$, $L_{3,z}$,
$L_{4,z}$)-type channel W state is more robust than GHZ state when the noisy
paramter ($\kappa$) is comparatively small while GHZ state becomes more robust
when $\kappa$ is large. In isotropic noisy channel we found that both states
preserve equal amount of quantum information. A relation between the average
fidelity and entanglement for the mixed state quantum channels are discussed.
