Crossed Cannizzaro products

I have understood the basic mechanism of cannizzaro reaction. But I am unable to understand the products formed in case of a crossed Cannizzaro reaction.

Suppose Benzaldehyde and Formaldehyde are heated in presence of $\ce{NaOH}$ to carry out Cannizzaro transformation. I have read that it is Benzyl Alcohol and Formic Acid formed in the greatest proportion and not Benzoic acid and Methyl alcohol.

Why is this so? How can we understand this in terms of the stability of Cannnizzaro intermediates?

The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.
\begin{aligned} &\ \ \ \ \ \ce{O} && \ \ \ \ \ \ce{O-} \\ &\ \ \ \ \ \parallel && \ \ \ \ \ \mid \\ &\ce{R-C-H +OH- ->} && \ce{R-C-H} \\ & \ && \ \ \ \ \ \mid \\ & \ && \ \ \ \ \ \ce{OH} \end{aligned}
Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:
• Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.