Answer
$P(I~win)=\frac{1}{962,598}\approx0.000001039$
Work Step by Step
The order in which the numbers are selected does not matter and no number can be selected more than once.
The sample space: all the combinations of 43 distinct numbers (from 1 to 43) taken 5 at a time:
$N(S)=_{43}C_5=\frac{43!}{5!(43-5)!}=\frac{43!}{5!\times38!}$
But, $43!=43\times42\times41\times40\times39\times(38\times37\times36\times...\times3\times2\times1)=43\times42\times41\times40\times39\times38!$
$N(S)=_{43}C_5=\frac{43\times42\times41\times40\times39\times38!}{5!\times38!}=\frac{43\times42\times41\times40\times39}{5\times4\times3\times2\times1}=\frac{115,511,760}{120}=962,598$
With one ticket: $N(I~win)=1$
Using the Classical Method (page 259):
$P(I~win)=\frac{N(I~win)}{N(S)}=\frac{1}{962,598}\approx0.000001039$
