To find the probability of purchasing at least one winning ticket out of 6 tickets when the probability of winning on a single ticket is $0.07$, we use the "at least once" rule.

Solution

1. Define the probability of not winning on a single ticket:
   If the probability of winning on a single ticket is $0.07$, then the probability of not winning on a single ticket is: $1 - 0.07 = 0.93$

2. Find the probability of not winning on all 6 tickets:
   If we have 6 tickets and the probability of not winning on each ticket is $0.93$, then the probability of not winning on all 6 tickets is: $(0.93)^6$

3. Calculate $(0.93)^6$: $(0.93)^6 \approx 0.665$

4. Apply the "at least once" rule:
   The probability of winning on at least one ticket (i.e., the complement of not winning on any ticket) is: $1 - (0.93)^6 \approx 1 - 0.665 = 0.335$

Answer

The probability of purchasing at least one winning lottery ticket out of 6 tickets is approximately: $\boxed{0.335} \text{ or } 33.5\%$

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Related Questions:

1. What is the probability of winning on at least two tickets out of six?
2. How does the probability change if the winning chance on a single ticket increases to 0.1?
3. What would the probability be if 10 tickets are purchased instead of 6?
4. Can you calculate the probability of winning on exactly one ticket out of 6?
5. How do we calculate the probability of winning at least once if the probability per ticket varies?

Tip:

For problems involving "at least one" scenarios, calculating the probability of the complementary event (none occurring) can simplify the process.