Q40
2 marksSection E

(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?

Probability
Probability — Classical Probability of Winning Prizes
Official Answer

Total tickets sold = 1,00,000. Winning tickets: 1 (first prize)+2 (second prize)+3 (third prize)=61 \text{ (first prize)} + 2 \text{ (second prize)} + 3 \text{ (third prize)} = 6.


Part (a): P(wins at least Rs 2,00,000)

  • "At least Rs 2,00,000" means winning the first prize (Rs 3,00,000) or a second prize (Rs 2,00,000): favourable tickets = 1+2=31+2 = 3.
  • P=3/1,00,000=P = 3/1,00,000 = 3/1000003/100000.

Part (b): P(does not win any amount)

  • P(wins something)=6/1,00,000=3/50,000P(\text{wins something}) = 6/1,00,000 = 3/50,000.
  • P(wins nothing)=1−3/50,000=P(\text{wins nothing}) = 1 - 3/50,000 = 49997/5000049997/50000.
classical probabilitycomplement rulefavourable outcomesat least probabilityjackpot draw1,00,000 tickets

Marking Scheme

  • 1Part (a) — 1 mark: correctly identifying favourable tickets = 1(first)+2(second)=31(\text{first})+2(\text{second}) = 3.
  • 2Part (a) — 1 mark: correct probability 3/1,00,0003/1,00,000.
  • 3Part (b) — 1 mark: correctly finding P(wins something)=6/1,00,000=3/50,000P(\text{wins something}) = 6/1,00,000 = 3/50,000.
  • 4Part (b) — 1 mark: correct final probability 1−3/50,000=49997/500001 - 3/50,000 = 49997/50000.

Hint

Use classical probability = favourable/total\text{favourable}/\text{total} (1,00,000 tickets); for 'at least Rs 2,00,000' add first+second prize tickets; for 'no win' use the complement rule 1−P(wins something)1 - P(\text{wins something}).

Quick Oral Answer

P(wins at least Rs 2,00,000)=3/1,00,000P(\text{wins at least Rs } 2,00,000) = 3/1,00,000, since only the 1 first-prize and 2 second-prize tickets qualify; P(wins nothing)=1−6/1,00,000=49997/50000P(\text{wins nothing}) = 1 - 6/1,00,000 = 49997/50000 by the complement rule.

Analysis & Explanation

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


Concept

  • Classical probability: P(event)=number of favourable outcomestotal number of equally likely outcomesP(\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(not A)=1−P(A)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(wins something)P(\text{wins something}) incorrectly by using only one prize category instead of summing all three (1+2+3=61+2+3=6).

Real-world relevance

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

Common Mistakes

  1. 1Wrongly including the Rs 50,000 third prize while computing 'at least Rs 2,00,000'.
  2. 2Forgetting to add all three prize-ticket counts (1+2+3=6)(1+2+3=6) when computing P(wins something)P(\text{wins something}) for the complement calculation.
  3. 3Not reducing the fraction 6/1,00,0006/1,00,000 to its simplest form 3/50,0003/50,000.

Interesting Facts

The overall probability of winning any prize here is just 6 in 1,00,000 (0.006%) — far lower than winning nothing, illustrating why real lotteries are structured with a very high 'house edge' relative to individual ticket buyers.

The complement rule P(A′)=1−P(A)P(A')=1-P(A) is one of Kolmogorov's foundational probability axioms (1933), still the basis of all modern probability theory taught in this chapter.

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Frequently Asked Questions

Why does 'at least Rs 2,00,000' exclude the third prize?

Because Rs 50,000 is less than Rs 2,00,000, so only the first prize (Rs 3,00,000) and second prizes (Rs 2,00,000) satisfy the 'at least' condition.

Is there a faster way to find P(wins nothing)P(\text{wins nothing}) without listing every non-winning scenario?

Yes — using the complement rule, P(wins nothing)=1−P(wins something)P(\text{wins nothing}) = 1 - P(\text{wins something}), where P(wins something)P(\text{wins something}) is simply (total winning tickets)/(total tickets)=6/1,00,000(\text{total winning tickets})/(\text{total tickets}) = 6/1,00,000.