#### Answer

$\displaystyle \frac{83}{84}$

#### Work Step by Step

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}$
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${}_{5}P_{3}=\displaystyle \frac{5!}{(5-3)!}=5\times 4\times 3$
${}_{10}P_{4}=\displaystyle \frac{10!}{(10-4)!}=10\times 9\times 8\times 7$
$1-\displaystyle \frac{ {}_{5}P_{3}}{{}_{10}P_{4}}=1-\frac{5\times 4\times 3}{10\times 9\times 8\times 7}$
$=1-\dfrac{1}{3\times 4\times 7}$
$=1-\displaystyle \frac{1}{84}=\frac{83}{84}$