(a) What is the probability that the person wins atleast Rs 2,00,000? OR (b) What is the probability that the person does not win any amount?

This part applies classical probability (favourable outcomes ÷ total outcomes) and the complement rule to a real prize-draw scenario.

Concept

• Classical probability: $P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}$ — here, each of the 1,00,000 tickets is equally likely to be Rohan's.
• "At least Rs 2,00,000" is the union of two disjoint winning categories (first and second prize), so their counts simply add.
• The complement rule $P(\text{not } A) = 1 - P(A)$ is the fastest way to compute "wins nothing" from "wins something."

Exam trap

• In part (a), a common mistake is including the third prize (Rs 50,000) in "at least Rs 2,00,000," which is incorrect since Rs 50,000 < Rs 2,00,000.
• In part (b), students sometimes compute $P(\text{wins something})$ incorrectly by using only one prize category instead of summing all three ($1+2+3=6$).

Real-world relevance

• This complement-rule shortcut (finding $P(\text{no prize})$ via $1 - P(\text{some prize})$) is exactly how real lottery and insurance companies quickly estimate payout probabilities without enumerating every losing scenario individually.