Question

Lucy gets home at a random time between 5:27 and 6:32 each evening, and watches the next news show that’s on. Network 1 airs news shows every half-hour (at 5:00, 5:30, 6:00, etc.); network 2 airs news shows at 5 minutes past each half-hour. Lucy will not start watching a show that’s already in progress.

(a) Let Ni be the event that Lucy ends up watching the news on network # i tomorrow, for i ∈ {1, 2}. Find P(Ni).

b) Find the expected amount of time that Lucy will have to wait to watch the news tomorrow.

Answer 1

(a) Lucy will watch on Network#1 and Network#2 with following probabilities between 5:27 to 6:32

Lucy Channel Watching Summary-Count and % between 5:27 to 6:32

P(Ni) where i=1 or 2	Count	%
P(N1)	54	0.82
P(N2)	12	0.18
Total	66	1.00

From above table,Probability of watching Network#1 by Lucy- P(N1)=0.82 and Probability of watching Network#2 by Lucy - P(N2) = 0.18

Detail Excel Sheet calculation for above is given below :-

(b) Average Waiting Time in Mins by Lucy for Tomorrow Watching is 9.59 Mins ie

Sum(of all Waiting Time-WT between 5:27 to 6:32)/Total N = (211+115+279+28)/66=9.59 Mins

Notation->AT->Arrival Time,Ni->Network Channel i (i=1 or 2),WT- Waiting Time in Mins for each AT