Question

What are the possible amounts, the person can win ?

Answer 1

Concept:

• The possible outcomes for a single ticket holder correspond to the various prize tiers available or winning no prize at all.

Step 1: List the available prize amounts
From the problem description, the different prize amounts are:

• First Prize: Rs 3,00,000
• Second Prize: Rs 2,00,000
• Third Prize: Rs 50,000

Step 2: Account for the possibility of not winning
Since there are only \( 1 + 2 + 3 = 6 \) winning tickets out of 1,00,000, it is possible for a person to not win any prize. In this case, the amount won is Rs 0.

Step 3: Combine all possible amounts
The set of possible winning amounts is:
\[ \{Rs 3,00,000, Rs 2,00,000, Rs 50,000, Rs 0\} \]

Answer 2

Concept:

• "At least Rs 2,00,000" means the person wins either a first prize (Rs 3,00,000) or a second prize (Rs 2,00,000).
• Probability \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Step 1: Identify the total number of outcomes
The total number of tickets sold is:
\[ N = 1,00,000 \]

Step 2: Identify the number of favorable outcomes
Number of first prize tickets = 1. Number of second prize tickets = 2. Total winning tickets for at least Rs 2,00,000: \[ n = 1 + 2 = 3 \]

Step 3: Calculate the probability
\[ P(\text{at least Rs 2,00,000}) = \frac{n}{N} \] \[ P = \frac{3}{1,00,000} \] \[ P = 0.00003 \]

Answer 3

Concept:

• The event "not winning any amount" is the complement of "winning any prize".
• \( P(\text{Not Win}) = 1 - P(\text{Win any prize}) \).

Step 1: Calculate the total number of winning tickets
Number of 1st prizes = 1. Number of 2nd prizes = 2. Number of 3rd prizes = 3. Total winning tickets: \[ \text{Win} = 1 + 2 + 3 = 6 \]

Step 2: Calculate the number of non-winning tickets
Total tickets = 1,00,000. Non-winning tickets: \[ \text{Non-Win} = 1,00,000 - 6 = 99,994 \]

Step 3: Calculate the probability
\[ P(\text{Not Win}) = \frac{99,994}{1,00,000} \] \[ P = 0.99994 \] Alternatively, using the complement: \[ P = 1 - \frac{6}{1,00,000} = 1 - 0.00006 = 0.99994 \]

Answer 4

Concept:
• This involves two independent events, \( A \) (winning any prize in the first jackpot) and \( B \) (winning the second jackpot).
• Probability of exactly one = \( P(A \cap B') + P(A' \cap B) \).
• Since events are independent, \( P(A \cap B') = P(A) \times P(B') \).

Step 1: Find individual probabilities of winning
For Jackpot 1 (event \( A \)): Total winning tickets = 6. \[ P(A) = \frac{6}{1,00,000} \] For Jackpot 2 (event \( B \)): \[ P(B) = \frac{1}{1,00,000} \]

Step 2: Find probabilities of not winning
\[ P(A') = 1 - \frac{6}{1,00,000} = \frac{99,994}{1,00,000} \] \[ P(B') = 1 - \frac{1}{1,00,000} = \frac{99,999}{1,00,000} \]

Step 3: Apply the formula for exactly one success
\[ P(\text{Exactly one}) = P(A)P(B') + P(A')P(B) \] \[ P = \left( \frac{6}{10^5} \times \frac{99,999}{10^5} \right) + \left( \frac{99,994}{10^5} \times \frac{1}{10^5} \right) \] \[ P = \frac{599,994 + 99,994}{10^{10}} = \frac{699,988}{10^{10}} \] \[ P \approx \frac{7}{10^5} = 0.00007 \text{ (approximately)} \]