Choosing exactly 2 damaged pieces (Probability)

Assume that each of the $4$ luggage items damaged by the handler was chosen independently and with equal probability. Let the number of insured items in the lot of $27$ be $i$ and uninsured $n$.

The probability that the handler damages $4$ items, only one of which is insured, is $$ \frac{\displaystyle {i \choose 1}{n\choose 3}}{\displaystyle{27\choose 4}}\,. $$

The probability that the handler damages $4$ items, none of which is insured, is $$ \frac{\displaystyle {n\choose 4}}{\displaystyle{27\choose 4}}\,. $$

We are told that the relationship between these probabilities is $$ \frac{\displaystyle {i \choose 1}{n\choose 3}}{\displaystyle{27\choose 4}}{}={}2\frac{\displaystyle {n\choose 4}}{\displaystyle{27\choose 4}}\,. $$

Simplifying, this implies that $$ n-2i{}={}3\,. $$ But, recall, that we also know $$ i{}+{}n=27\,. $$ Thus, solving this system of equations yields $i{}={}8$ and $n{}={}19$, from which the probability that exactly $2$ of the $4$ damaged items are insured may be calculated as $$ \frac{\displaystyle {8 \choose 2}{19\choose 2}}{\displaystyle{27\choose 4}}{}\approx{}0.2728\,. $$