[[V]]

I would think the order goes as : I, II, III
While actually, the answer is II, I, III

Here is my reasoning for my answer:
The carbon where the chlorine was added, originally had a radical prior to the chlorination.
So the stability of the radical determines the amount of product. Clearly compound I had the most stable radical (2' carbon)
Then II > III because in II the beta-carbon is more substituted and delivers more sigma induction. Also, there are 2 identical methyl groups that could have been chlorinated, so statistically this one is more likely.

I dont understand why II comes first in the answer. Does three 1' carbons outweigh one 2' carbon?

Thanks for reading.

[fledarmus]

The difference between primary, secondary, and tertiary radicals is large compared to the difference between a  primary radical adjacent to a tetraalkyl carbon and a primary radical adjacent to a dialkyl carbon. If the only thing you have to distinguish two radicals is the very small effect of an additional alkyl group on a beta-carbon, the effect of the statistical likelihood of one of 9 identical protons being pulled compared to one of 3 is stronger.

If the stabilities of the two different primary groups were absolutely identical you would expect statistically a 3:1 mixture of products - you would probably see a smaller ratio due to the influence you mention, but not enough to change which is the predominant product.

[orgopete]

I think this is a calculation problem, not a predict the major product. See the example posted here:
http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=324#p324