Probability Calculation: At Least One Trip Caught

Understanding the Problem

We need to find the probability that a ticketless traveler is caught during at least one of their 4 trips. We are given the probability of being caught on a single trip.

Key Information

• Probability of being caught on one trip: $P(\text{Caught}) = 0.1$
• Number of trips: $n = 4$
• We need $P(\text{Caught on at least one trip})$

Applying Complementary Probability

Calculating the probability of being caught on "at least one" trip is often simpler using the complement rule. The complement of being caught on "at least one" trip is being caught on "none" of the trips.

$ P(\text{At least one caught}) = 1 - P(\text{No trips caught}) $

Calculating Probability of Not Being Caught

First, find the probability of *not* being caught on a single trip:

$ P(\text{Not Caught}) = 1 - P(\text{Caught}) = 1 - 0.1 = 0.9 $

Calculating Probability of Not Being Caught on Any Trip

Since the trips are independent events, the probability of not being caught on any of the 4 trips is:

$ P(\text{No trips caught}) = P(\text{Not Caught on Trip 1}) \times P(\text{Not Caught on Trip 2}) \times P(\text{Not Caught on Trip 3}) \times P(\text{Not Caught on Trip 4}) $

$ P(\text{No trips caught}) = (0.9) \times (0.9) \times (0.9) \times (0.9) = (0.9)^4 $

Final Probability Calculation

Now, substitute this back into the complementary probability formula:

$ P(\text{At least one caught}) = 1 - P(\text{No trips caught}) = 1 - (0.9)^4 $

Conclusion

The probability that the traveler will be caught during at least one of the 4 trips is $1 - (0.9)^4$. This matches Option A.