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:

1. What are the possible amounts the person can win?
2. 1. What is the probability that the person wins atleast ₹ 2,00,000?
      OR
   2. What is the probability that the person does not win any amount?
3. 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.

Answer 1

(i)

A person who buys a ticket can win any one of the designated prizes. The possible amounts are:

• ₹ 3,00,000 (First Prize)
• ₹ 2,00,000 (Second Prize)
• ₹ 50,000 (Third Prize)

(Note: They can also “win” ₹ 0 if they do not hold a winning ticket.)

(ii) (a)

Total number of tickets sold (n) = 1,00,000

Winning “at least ₹2,00,000” means winning either the first prize or one of the second prizes.

• Number of 1st prize tickets = 1
• Number of 2nd prize tickets = 2
• Total favourable tickets = 1 + 2 = 3

∴ P(wins atleast 2,00,000) = `3/(1,00,000)`, or P = 0.00003

OR

(ii) (b)

To find the probability of not winning, we first find the total number of winning tickets.

• Total winning tickets = 1 (1st) + 2 (2nd) + 3 (3rd) = 6
• Total non-winning tickets = 1,00,000 − 6 = 99,994

∴ P(no win) = `(99,994)/(1,00,000)` or P = 0.99994

(iii)

In this scenario, Rohan has two tickets for two different jackpots.

• Event A: Wins Jackpot 1 (1st prize). P(A) = `1/(1,00,000)`
• Event B: Wins Jackpot 2. P(B) = `1/(1,00,000)`

To win on exactly one ticket, he must win the first and lose the second, OR lose the first and win the second:

P(Exactly one) = P(A) × P(B′) + P(A′) × P(B)

P(Exactly one) = `(1/10^5 xx (99,999)/10^5) + ((99,999)/10^5 xx 1/10^5)`

P(Exactly one) = `(99,999 + 99,999)/(10,00,00,00,000)`

P(Exactly one) = `(1,99,998)/10^10`

∴ P(Exactly one) ≈ 0.0000199998