Question

For my flight home to Germany this summer, I’ll have a stop at Reykjavk airport, Iceland, for 55 minutes. Assuming probabilities of waiting time are evenly distributed (i.e. no delays) calculate the following probabilities:

i) What is the probability I’ll only get the very last minute, 55th minute?

ii) How likely is it that I have to wait for more than half an hour?

Answer 1

Consider X denotes waiting time.

Since waiting time are evenly distributed.

i.e. The distribution of random variable X is discrete Uniform.

The probability mass function of random variable X is

i) Required Probability = P ( X =55)

= 1/ 55

= 0.0182.

ii) Required Probability = P ( X > 30)

= 0.4545

Answer 2

To calculate the probabilities in this scenario, we assume that the waiting time at Reykjavik airport is evenly distributed with no delays. Let's calculate the probabilities:

i) Probability of getting the very last minute, 55th minute: Since the waiting time is evenly distributed, the probability of getting the very last minute is equal to the probability of getting any specific minute during the 55-minute duration. Therefore, the probability of getting the 55th minute is 1/55.

ii) Probability of waiting for more than half an hour: To calculate this probability, we need to determine the proportion of the 55-minute duration that represents more than half an hour. Half an hour is equivalent to 30 minutes, so we need to find the probability of getting a waiting time greater than 30 minutes.

Since the waiting time is evenly distributed, the probability of waiting for more than half an hour is equal to the proportion of the remaining time after 30 minutes. The remaining time is 55 minutes - 30 minutes = 25 minutes. Therefore, the probability of waiting for more than half an hour is 25/55.

To summarize: i) The probability of getting the very last minute, 55th minute, is 1/55. ii) The probability of waiting for more than half an hour is 25/55 or approximately 0.4545.