I need help with this problem,

To get to work, a commuter must cross train tracks. The time the train arrives varies slightly from day to day, but the commuter estimates he'll get stopped approximately 10% of the time on his way to work. For each problem below, assume that getting stopped by the train on one day is independent of getting stopped on any other day.


c) During a 4-day work week, what is the probability he gets stopped exactly once?
I think the answer is 1-0.90^4 which equal 0.3439 is that the correct answer?

d) During a 4-day work week, what is the probability he gets stopped at most once?

I am really unsure about this question, but my best guess would be 1 minus the answer from section C so 1-0.3439 which is 0.6561. Is this correct or am I way off.

c) No. To be stopped once means he is not stopped 3 times. The probability of that happening is C(4,1)(0.1)^1 (0.9)^3. The combination term is in this because the day he is stopped could be any one of the 4 days.

What you calculated with 1 - 0.9^4 is the probability of being stopped at least once. In other words it also includes the probability of being stopped 2, 3, or 4 times as well.

d) The probability of being stopped at most once is equal to the probability of being stiopped precisely once (which we just calculated) plus the probability of being stopped 0 times. Can you complete this now?

Thank You! I got the answer now