Tom often waits for the 192 bus at the University of Queensland's Lake Bus Station, where he has noticed that it departs from the Lake Bus at a rate of six buses per hour. What is the probability that he will be able to wait for one bus in a 1-minute period (round your answer to four decimal places)?
Question
Tom often waits for the 192 bus at the University of Queensland's Lake Bus Station, where he has noticed that it departs from the Lake Bus at a rate of six buses per hour. What is the probability that he will be able to wait for one bus in a 1-minute period (round your answer to four decimal places)?
Solution
This is a Poisson distribution problem. The Poisson distribution is used to model the number of events occurring within a given time period.
The formula for the Poisson probability is:
P(X=k) = (λ^k * e^-λ) / k!
where:
- P(X=k) is the probability of k events in the interval
- λ is the average rate of value
- e is the base of the natural logarithm, approximately equal to 2.71828
- k is the actual number of successes
In this case:
- λ = 6 buses per hour / 60 minutes per hour = 0.1 buses per minute
- k = 1 (we want the probability of one bus arriving in a 1-minute period)
Substituting these values into the formula, we get:
P(X=1) = (0.1^1 * e^-0.1) / 1!
Calculating the above expression gives a probability of approximately 0.0905 when rounded to four decimal places.
So, the probability that he will be able to wait for one bus in a 1-minute period is 0.0905.
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