問題解答
HKAL 95II Qn11
Every morning, Mr. Chan has to reach his officer by 9
a.m. His office in a 60-storey building which is served
by four lifts A, B, C and D. Lifts A and B stop at any
floor between the ground and the 30th floors. To go to
the upper part of the building, one has to go to the 30th
floor first and then take lift C or D, both of which stop
at any floor between the 30th and the 60th floors. The
four lifts work independently. On any day, the
probability that lift A is out of service before 9 a.m.
is 0.1, 0.2 and 0.2 respectively.
(a) In a year of 365 days, what would be the expected
number of days on which only one lift id out of service
before 9 a.m.?
(3 marks)
(b) Given that Mr. Chan's office is on the 25th floor,
find the probability that he can reach his office
punctually by lift on a particular day.
(2 marks)
(c) Now Mr. Chan is transferred to another office on the
45th floor.
(i) Find the probability that he can reach his office
punctually by lift on a particular day.
(ii) Find the probability that out of 20 days, he can
reach his office punctually by lift on at least 18 days.
(iii) Given that two lifts are out of service before 9
a.m. on a particular day, find the probability that he
can still reach his office punctually by lift.
(10 marks)
Solution:
(a) probability that only one lift is out of service
before 9:00 am
= (0.1*0.9*0.8*0.8+0.9*0.9*0.2*0.8)*2
= 0.3744
i.e. expected number of days = 365*0.3744 = 136.66 =137
(b) probability = 1-0.1*0.1 = 0.99
(c) (i) probability = (1-0.1*0.1)*(1-0.2*0.2) =0.9504
(ii) 20C18(0.9504)18(0.0496)2+20C19(0.9504)19(0.0496)+0.950420
= 0.9259
(iii) P(reach punctually │2 lifts are out of service)
=[(0.1*0.9*0.2*0.8)*4]/[0.1*0.1*0.8*0.8+0.9*0.9*0.2*0.2
+(0.1*0.9*0.2*0.8)*4]
= 0.5975