The probability that a ticketless traveller is caught during a rip is 0.1. If the traveller makes 4 trips, the probability that he/she will be caught during at least one of the trips is:
1‐(0.9)4
Probability of Being Caught
Let's analyze the problem about the ticketless traveller and the probability of being caught during their trips.
We are given the following information:
- The probability that a ticketless traveller is caught during a single trip is 0.1.
- The traveller makes 4 independent trips.
We need to find the probability that the traveller will be caught during at least one of the trips.
Probability of Not Being Caught
First, let's determine the probability that the traveller is not caught during a single trip. If the probability of being caught is $P(\text{Caught}) = 0.1$, then the probability of not being caught is the complement of this event.
Probability of not being caught on a single trip, $P(\text{Not Caught}) = 1 - P(\text{Caught})$
$\qquad P(\text{Not Caught}) = 1 - 0.1 = 0.9$
Probability Over Multiple Independent Trips
The traveller makes 4 trips, and we assume these trips are independent events. This means the outcome of one trip does not affect the outcome of any other trip.
To find the probability that the traveller is not caught during any of the 4 trips, we multiply the probabilities of not being caught on each individual trip:
$P(\text{Not Caught on any of the 4 trips}) = P(\text{Not Caught on trip 1}) \times P(\text{Not Caught on trip 2}) \times P(\text{Not Caught on trip 3}) \times P(\text{Not Caught on trip 4})$
$\qquad P(\text{Not Caught on any of the 4 trips}) = (0.9) \times (0.9) \times (0.9) \times (0.9)$
$\qquad P(\text{Not Caught on any of the 4 trips}) = (0.9)^4$
Probability of Being Caught at Least Once
The event "being caught during at least one of the trips" is the complement of the event "not being caught during any of the trips".
So, the probability of being caught during at least one trip is:
$P(\text{Caught at least once}) = 1 - P(\text{Not Caught on any of the 4 trips})$
$\qquad P(\text{Caught at least once}) = 1 - (0.9)^4$
Comparing with Options
Let's look at the given options:
- 1‐(0.9)4
- (1‐0.9)4
- 1‐(1‐0.9)4
- (0.9)4
Our calculated probability is $1 - (0.9)^4$. This matches the first option.
The second option, $(1-0.9)^4 = (0.1)^4$, would be the probability of being caught on every single trip.
The fourth option, $(0.9)^4$, is the probability of not being caught on any trip.
The third option is $1-(1-0.9)^4 = 1-(0.1)^4$, which is $1$ minus the probability of being caught on every trip.
Conclusion
The probability that the ticketless traveller is caught during at least one of the 4 trips is $1 - (0.9)^4$.
Important Questions from Probability
Three dice are thrown. What is the probability of getting a sum which is a perfect square?
Two distinct natural numbers from 1 to 9 are picked at random. What is the probability that their product has 1 in its unit place?
Two dice are thrown. What is the probability that difference of numbers on them is 2 or 3 ?
Suppose that there is a chance for a newly constructed building to collapse, whether the design is faulty or not. The chance that the design is faulty is 10%. The chance that the building collapses is 95% if the design is faulty, otherwise it is 45%. If it is seen that the building has collapsed, then what is the probability that it is due to faulty design?
What is the probability that all three boys sit together?
