1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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 $\frac{162}{3125}$ $\endgroup$
    – Gono
    Aug 7, 2017 at 10:24
  • $\begingroup$ @Gono, that is not the right answer, it's 0.00246 $\endgroup$
    – J-Dorman
    Aug 7, 2017 at 10:38
  • $\begingroup$ 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 :-) $\endgroup$
    – Gono
    Aug 7, 2017 at 10:46
  • $\begingroup$ Well that is the answer thats published in the book, but it might be wrong @Gono $\endgroup$
    – J-Dorman
    Aug 7, 2017 at 10:48
  • $\begingroup$ @Gono Also, you are doing the number of days the bus is late, not that it is on time $\endgroup$
    – J-Dorman
    Aug 7, 2017 at 10:49

2 Answers 2

Reset to default
2
$\begingroup$

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 : $P(late).(P(on time))^4$

$\endgroup$
2
$\begingroup$

To get the "right" answer you need to use a seven day week. Expressing the probability of lateness as 2/5 can throw you off track. On-time probability on any given day is 0.6, and late 0.4. (0.6)(0.4)(0.4)(0.4)(0.4)(0.4)(0.4)=0.0024576

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .