Symbiosis National Aptitude Test SNAP 2012 Paper

No. of candidate for first post = 5
No. of candidate for 2nd post = 8
No. of candidate for 3rd post = 7
P (getting 1st post) = $\frac{1}{5}$, P (getting 2nd post) = $\frac{1}{8}$ and P (getting 3rd post) = $\frac{1}{7}$
∴ P (not getting 1st post) = $\frac{4}{5}$, P (not getting 2nd post ) = $\frac{7}{8}$ and P (not getting 3rd post) = $\frac{6}{7}$
= $1-\left(\frac{4}{5}\times \frac{7}{8}\times \frac{6}{7}\right)$
∴ P (getting at least one pest) = 1 - $\frac{3}{5}$ = $\frac{2}{5}$