The hydrolyzed product of an ester, when reduced by $LiAlH_4$, gives:

The reaction proceeds in two conceptual steps based on the question wording:

1. Hydrolysis of Ester: Esters ($RCOOR'$) undergo hydrolysis to form a carboxylic acid ($RCOOH$) and an alcohol ($R'OH$) [1, 2].
2. Reduction by $LiAlH_4$: Lithium aluminium hydride ($LiAlH_4$) is a strong reducing agent. It reduces carboxylic acids ($RCOOH$) to primary alcohols ($RCH_2OH$). The alcohol ($R'OH$) produced in the first step remains an alcohol.

Alternatively, direct reduction of an ester by $LiAlH_4$ also cleaves the ester bond to yield two alcohols: one corresponding to the acyl group (reduced to a primary alcohol) and one corresponding to the alkoxy group. In either interpretation, the final product mixture consists of two alcohols.