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.
Important Questions from Discrete Probability
- A biased six-faced die is tossed once. Suppose that the probability of any prime number showing up is twice that of any non-prime number showing up. Then, the probability that an odd number will show up is
- Let $X$ and $Y$ be independent Poisson random variables with means $4$ and $2$, respectively. Then, which of the following statements are true?
- Consider the M/M/1 queue in which customers arrive according to a Poisson process with rate $3$ and successive service times are independent exponential random variables having mean $\frac{1}{9}$. Let $P_n$ be the long run probability that there are exactly $n$ customers in the system. Then, which of the following statements are true?
- Let $X$ be a Binomial$(n, p)$ random variable, where $n \in \{5,6\}$ and $p\in \{\frac{1}{4}, \frac{3}{4}\}$. If $X = 3$ is observed, then the maximum likelihood estimate of $(n, p)$ is
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
