CBSE Board Examination 2026 · Section E · 4 marks

Q37: Case study: an online jackpot

Board 2026Ch 13 · Probability

The question

In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each.

A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold there by raising a fund of ₹ 1,00,00,000.

Rohan bought one ticket.

Based on given information, answer the following questions :

(i) What are the possible amounts, the person can win? (1 mark)

(ii) (a) What is the probability that the person wins atleast ₹ 2,00,000? (2 marks)

(ii) (b) What is the probability that the person does not win any amount? (2 marks)

(iii) In another jackpot, Rohan also bought a ticket having a prize money of ₹ 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot. (1 mark)

Exam answer

(i) The possible amounts are ₹ 3,00,000, ₹ 2,00,000 and ₹ 50,000.

Why?

There are three kinds of prize. One ticket can win at most one prize, so the amounts a winner can get are exactly the three prize values.

(ii) (a) Total number of tickets $= 100000$ ⁠. He wins at least ₹ 2,00,000 if his ticket gets the first prize or a second prize: $1 + 2 = 3$ tickets.

Why?

"At least ₹ 2,00,000" means ₹ 2,00,000 or more: the first prize (₹ 3,00,000, 1 ticket) or a second prize (₹ 2,00,000, 2 tickets).

Watch out: the third prizes of ₹ 50,000 do not count.

Required probability $= \frac{3}{100000}$ ⁠.

Why?

Every ticket is equally likely to be drawn, so

P = \frac{\text{favourable tickets}}{\text{all tickets}}

(ii) (b) Number of prize tickets $= 1 + 2 + 3 = 6$ ⁠.

\begin{aligned} P(\text{no prize}) &= 1 - \frac{6}{100000} \\ &= \frac{99994}{100000} \end{aligned}

Why?

"No prize" is the complement of "some prize", so

P(\text{no prize}) = 1 - P(\text{prize})

$P(\text{prize}) = \frac{6}{100000}$ ⁠, since 6 of the 100000 tickets win.

Write 1 as $\frac{100000}{100000}$ to subtract:

\begin{aligned} 1 - \frac{6}{100000} &= \frac{100000}{100000} - \frac{6}{100000} \\ &= \frac{100000 - 6}{100000} \\ &= \frac{99994}{100000} \end{aligned}

Required probability $= \frac{99994}{100000} = \frac{49997}{50000}$ ⁠.

Why?

Divide the numerator and the denominator by 2 to simplify. Either form is correct.

\frac{99994}{100000} = \frac{99994 \div 2}{100000 \div 2} = \frac{49997}{50000}

(iii) Let $E$ ⁠: he wins on the first ticket (any of its 6 prizes), and $F$ ⁠: he wins the jackpot on the second ticket. $E$ and $F$ are independent.

Why?

The two tickets are in different jackpots, so the result of one does not affect the other. For independent events the probability of both is the product of the probabilities.

P(E) = \frac{6}{10^5},\quad P(F) = \frac{1}{10^5}

\begin{aligned} &P(\text{exactly one}) \\ &= P(E)P(F') + P(E')P(F) \\ &= \frac{6}{10^5}\cdot\frac{99999}{10^5} + \frac{99994}{10^5}\cdot\frac{1}{10^5} \\ &= \frac{599994 + 99994}{10^{10}} \end{aligned}

Why?

"Exactly one" happens in two separate ways: win on the first and not the second, or win on the second and not the first. These cannot both happen, so add them.

By the complement rule,

\begin{aligned} P(E') &= 1 - \frac{6}{10^5} = \frac{99994}{10^5} \\ P(F') &= 1 - \frac{1}{10^5} = \frac{99999}{10^5} \end{aligned}

Here $10^5 = 100000$ ⁠, so $1 = \frac{100000}{10^5}$ ⁠, and $100000 - 6 = 99994$ ⁠, $100000 - 1 = 99999$ ⁠.

Multiply tops together and bottoms together. On the bottom, $10^5 \times 10^5 = 10^{5 + 5} = 10^{10}$ ⁠:

\begin{aligned} \frac{6}{10^5}\cdot\frac{99999}{10^5} &= \frac{6 \times 99999}{10^{10}} = \frac{599994}{10^{10}} \\ \frac{99994}{10^5}\cdot\frac{1}{10^5} &= \frac{99994 \times 1}{10^{10}} = \frac{99994}{10^{10}} \end{aligned}

Then $6 \times 99999 = 599994$ ⁠, and $599994 + 99994 = 699988$ ⁠. Dividing the numerator and the denominator by 4 gives the simpler form in the answer.

\begin{aligned} \frac{699988}{10^{10}} &= \frac{699988 \div 4}{10000000000 \div 4} \\ &= \frac{174997}{2500000000} \end{aligned}

$\therefore$ Required probability $= \dfrac{699988}{10^{10}}$ $= \dfrac{174997}{2500000000}$ ⁠.

Where this comes from

Chapter 13, Probability: Probability: lottery prizes, independent events.

Question type: (iii) At least one, or exactly one

How to spot it: click to see

(iii) Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: The two tickets are independent, and the question says "exactly one". Use $P(E)P(F') + P(E')P(F)$ with $P(E) = \frac{6}{10^5}$ and $P(F) = \frac{1}{10^5}$ ⁠.

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