To solve this problem using the "at least once" rule, we calculate the probability of the complement event (not winning any tickets) and subtract it from 1.

Definitions:

• $p = 0.06$: Probability of winning on a single ticket.
• $q = 1 - p = 0.94$: Probability of not winning on a single ticket.
• $n = 9$: Number of tickets purchased.

Complement Probability:

The probability of not winning any tickets is: $P(\text{not winning any}) = q^n = 0.94^9$

Complement Probability Calculation:

$P(\text{not winning any}) = 0.94^9 \approx 0.5678$

At Least Once Probability:

The probability of winning at least one ticket is: $P(\text{at least one win}) = 1 - P(\text{not winning any})$ $P(\text{at least one win}) = 1 - 0.5678 = 0.4322$

Final Answer:

The probability of purchasing at least one winning ticket out of 9 tickets is approximately: $P(\text{at least one win}) \approx 0.4322 \text{ or } 43.22\%.$

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Would you like a detailed breakdown of the calculation? Here are some follow-up questions:

1. How does this result change if the number of tickets increases or decreases?
2. What if the winning probability per ticket changes?
3. How can this concept be extended to other events beyond lotteries?
4. Can we calculate the expected number of wins given this scenario?
5. How is this different from calculating probabilities for independent events occurring?

Tip: Always calculate the complement first when solving "at least once" problems—it simplifies the math significantly!