# Discrete Random Variables - Binomial Distribution

The 8 am bus will run late on average 2 days out of 5. For any week of the year at random, find the probability of the bus being on time only on Monday.

I am totally stuck on this question. For me, I thought it would just be the probability of the bus being on time multiplied by 1/7 since it is a seven day week. But, that isn't the answer.

• First: "on average 2 days out of 5" means that you have to consider days monday till friday. "on time only on monday" means: "On time on monday, but not on time on tuesday, wednesday, thursday, friday" or "on time on just ONE day of the 5 day week AND this ONE day is monday". Does that help? For checking your result: The correct answer is $\frac{162}{3125}$ – Gono Aug 7 '17 at 10:24
• @Gono, that is not the right answer, it's 0.00246 – J-Dorman Aug 7 '17 at 10:38
• Where does your solution come from? Mine is, according to your question, the result of the following: Let $X \sim Bin(5,\frac{2}{5})$ the number of days the bus is late. Then $$P(X=1) = {5\choose 1}\cdot \frac{2}{5}\cdot \left(\frac{3}{5}\right)^{4} = 2 \left(\frac{3}{5}\right)^{4}$$ is the probability of being late on one day. If it should be the monday we have a chance of $$\frac{1}{5} \cdot 2 \cdot \left(\frac{3}{5}\right)^{4} = \frac{162}{3125}$$ edit: And now we got the mistake… the question asks for "being on time on monday" not "being late on only on monday" but you can adapt it :-) – Gono Aug 7 '17 at 10:46
• Well that is the answer thats published in the book, but it might be wrong @Gono – J-Dorman Aug 7 '17 at 10:48
• @Gono Also, you are doing the number of days the bus is late, not that it is on time – J-Dorman Aug 7 '17 at 10:49

You have $P(day1 = late, day2=on time, day3=on time, day4 = on time, day5=one time)$. The commas stand for AND, hence since the events are independant : $P(late).(P(on time))^4$