For independent opportunities, "at least one" is fastest via complements: \(1 - \prod (1 - p_i)\). It avoids messy inclusion-exclusion with many terms.
Step 1: Model each post as an independent fair choice among candidates.
Probability he gets the first post \(= \dfrac{1}{5}\); second \(= \dfrac{1}{8}\); third \(= \dfrac{1}{7}\).
We seek \(P(\text{at least one post})\).
Step 2: Use the complement (none of the posts).
\(P(\text{none}) = \left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{8}\right)\left(1 - \dfrac{1}{7}\right) = \dfrac{4}{5}\cdot\dfrac{7}{8}\cdot\dfrac{6}{7}\).
Cancel \(7\): \(= \dfrac{4}{5}\cdot\dfrac{6}{8} = \dfrac{4}{5}\cdot\dfrac{3}{4} = \dfrac{3}{5}\).
Step 3: Convert back to "at least one".
\(P(\text{at least one}) = 1 - P(\text{none}) = 1 - \dfrac{3}{5} = \boxed{\dfrac{2}{5}}\).
In a special racing event, the person who enclosed the maximum area would be the winner and would get ₹ 100 every square metre of area covered by him/her. Jonsson, who successfully completed the race and was the eventual winner, enclosed the area shown in the figure below. What is the prize money won?
\(\textit{Note: The arc from C to D makes a complete semi-circle. Given: }\) $AB=3$ m, $BC=10$ m, $CD=BE=2$ m.

A lawn is in the form of an isosceles triangle. The cost of turfing on it came to $₹ 1{,}200$ at ₹ 4 per m$^2$. If the base be 40 m long, find the length of each side.