CBSE Class 12 Mathematics 2026: question 37

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Step-by-step solution

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1. (i) Rohan can win ₹3,00,000 (first prize), ₹2,00,000 (second prize) or ₹50,000 (third prize).

   1 mark

2. (ii)(a) Winning at least ₹2,00,000 means winning the first prize or a second prize. There are $1+2=3$ such tickets out of $1,00,000$.
   So $P=\frac{3}{100000}$.

   2 marks

3. (ii)(b) There are $1+2+3=6$ winning tickets in all.
   $P(\text{no prize})=1-\frac{6}{100000}=\frac{99994}{100000}=\frac{49997}{50000}$.

   2 marks

4. (iii) In the first jackpot, $P(\text{win})=\frac{6}{100000}$ and $P(\text{lose})=\frac{99994}{100000}$. In the second jackpot, $P(\text{win})=\frac{1}{100000}$ and $P(\text{lose})=\frac{99999}{100000}$. The two draws are independent.

   $P(\text{exactly one win})=\frac{6}{10^5}\cdot \frac{99999}{10^5}+\frac{99994}{10^5}\cdot \frac{1}{10^5}$
   $=\frac{599994+99994}{10^{10}}=\frac{699988}{10^{10}}=\frac{174997}{2500000000}$.

   1 mark

Answer: (i) ₹3,00,000, ₹2,00,000, ₹50,000; (ii)(a) $\frac{3}{100000}$; (ii)(b) $\frac{49997}{50000}$; (iii) $\frac{699988}{10^{10}}=\frac{174997}{2500000000}$

Revise the chapter

• Chapter: Probability (notes, formulas, where marks are lost and every past question)