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
- 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?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Consider the problem of testing $H_0 : \theta = 1$ vs $H_1 : \theta = \frac{1}{2}$ where $\theta$ is the mean of a Poisson random variable. Let $X$ and $Y$ be a random sample from Poisson ($\theta$) distribution. Consider the following test procedure:
Reject $H_0$ if either $X = 0$ or $(X = 1 \text{ and } X + Y \leq 2)$; otherwise accept $H_0$.
Which of the following are true?
In a football league, the goals scored by home teams over 380 matches have the following frequency distribution.
Number of goals 0 1 2 3 4 5 Frequency 92 121 91 50 19 7 The average goals scored by home teams is 1.49. We want to test $H_0$: Goal distribution is Poisson. Based on observations the value of the $\chi^2$-statistic for goodness of fit is 1.27. Given $\chi^2_{0.05, 6} = 1.64, \chi^2_{0.05, 5} = 1.15, \chi^2_{0.95, 6} = 12.59$ and $\chi^2_{0.95, 5} = 11.07$, which of the following are true?
