1. Jessie needs to travel to school by bus and train. His bus is due to reach the interchange at 0745 hrs and is considered late if it reaches after that. The train he hopes to catch is due to leave the station at 0748 hr. The time of arrival of the bus at the interchange has a normal distribution with mean at 0744 hrs and standard deviation of 2 minutes. The departure time of the train has a nornal distribution with mean at0748 hrs and standard deviation1 minute.
(i)By defining X as the number of minutes after 0744 hrs in which the bus arrives, find the probability that the bus is late.
(ii) Show that the probability that the train departs before bus arrives is 0.0368
(iii) Find the probability that in a period of 5 days there are at least 3 days on which the train departs before the bus arrives.
State an assumption you made in calculating the probabilities in (ii) and (iii)
2 For any married couples in a club, the probability that the husband is the only child is 4/5. The probability that the wife is the only child is 1/2. The probability that the wife is the only child, given that the husband is not the only child is 2/3
2 married couples are chosen at random. Using a venn diagram, find probability that only one of the two husbands and only one of the 2 wives is the only child.
If possible, please do offer my a step-by-step working (I will only obtain the solutions after September, lol)
Thanks everyone.
Please post a meaningful reply if you meant to help. Thanks.
Hi Audi,
Q1:
(i) Bus is late if time > 0745h.
Given: A ~ N(0744, 2^2).
Required probability = P(A > 0745) = 0.309, by GC.
P.S. I do not see how the definition of X would help; it would confuse the student instead.
(ii) Given: D ~ N(0748, 1^2).
Consider D - A ~ N(4, 5).
Required probability = P(D - A < 0) = 0.03681, so 0.0368 is the required value.
(iii) Consider X ~ B(5, 0.03681), from which we want to find P(X >= 3).
Give an appropriate assumption for the Binomial distribution to be valid in the context of the question.
Thanks.
Cheers,
Wen Shih
Hi Audi,
Q2:
Let A be the event that the husband is the only child.
P(A) = 4/5.
Let B be the event that the wife is the only child.
P(B) = 1/2.
Given: P(B | A') = 2/3.
1. Find P(B intersect A').
2. Draw a Venn diagram (as suggested) and fill in all the values for
P(A intersect B'), P(A intersect B), P(A' intersect B').
3. To find the required probability, consider two cases. The first case is where one couple contains the only-child husband and the only-child wife. The second case is where one couple contains the only-child husband and the other couple contains the only-child wife.
Thanks.
Cheers,
Wen Shih