Answer: Given that the commuter gets stopped by a train with a probability of 13%, we can calculate the probabilities for different scenarios using the principles of probability.
Let's denote:
P(stop) as the probability of getting stopped by a train (0.13 in this case).
P(no stop) as the probability of not getting stopped by a train (1 - P(stop), which is 0.87 in this case).
For the following calculations, I'll assume that each day's events are independent of the others.
i. Probability of getting stopped on Monday and Tuesday:
Since the events are independent, the probability of both events occurring is the product of their individual probabilities.
P(getting stopped on Monday and Tuesday) = P(stop on Monday) * P(stop on Tuesday)
= P(stop) * P(stop)
= 0.13 * 0.13
= 0.0169
ii. Probability of not getting stopped on Monday, Tuesday, and Wednesday:
Again, since the events are independent, the probability of all three events not occurring is the product of their individual probabilities of not occurring.
P(not getting stopped on Monday, Tuesday, and Wednesday) = P(no stop on Monday) * P(no stop on Tuesday) * P(no stop on Wednesday)
= P(no stop) * P(no stop) * P(no stop)
= 0.87 * 0.87 * 0.87
≈ 0.64639
So, the answers are:
i. The probability that the commuter gets stopped on Monday and Tuesday is approximately 0.0169.
ii. The probability that the commuter does not get stopped on Monday, Tuesday, and Wednesday is approximately 0.64639.