Question

In the 1950s, there were only a few television channels. Assume each TV channel is independent, and they each play three 1-minute commercials each hour on average. There is no relationship between the schedules of each channel or the time of day/hour and whether or not a commercial is playing. Assume your TV has three channels, you hate commercials and you are an infinitely fast channel changer.

a) What is the probability that at any given moment, there are commercials playing on each channel (and therefore nothing to watch)?

b) What is the probability that there is at least one channel not playing a commercial (and therefore, at least something to watch)?

c) What is the probability that Channel 1 or Channel 2 has a commercial playing at any given time? Do not round your answer.

d) If I upgrade my antenna and can now receive a fourth channel, however, this channel is owned by large corporations who randomly advertise for fifteen minutes each hour, what is the probability that at any given moment there are commercials playing on each channel (and therefore nothing to watch)?

e) Now assume that I am back to only three channels. Advertisers have figured out that I just flip through channels to watch anything that is not a commercial. The channels are starting to work together to make sure advertisers are getting their money's worth. Now assume that they only begin commercials at each minute mark, and they have increased the likelihood of a 1-minute commercial occurring from the 29th to 30th minute of each hour. The new probability of a commercial during this interval is 0.6 for each channel.  

What is the probability that it is the 29th to 30th minute of the hour AND that I have nothing to watch (commercials on all three channels)? Do not round your answers.

Answer 1

given

There are three tv channels which are independent and each channel plays three 1 minute commercial each on average. Moreover there is no relationship between the schedules of each channel or the time of day/hour and whether or not a commercial is playing.

therefore probability for each channel that it will be playing a commercial = 3/60

suppose,

now

P(X_i=1)=1/20 for all i=1,2,3 and

P(X_i=0)=19/20 for all i=1,2,3

a.

the probability that at a given moment there are commercials playing at each channel

=P(X₁=1,X₂=1,X₃=1)

=P(X₁=1).P(X₂=1).P(X₃=1) [since independent]

=(1/20).(1/20).(1/20)

=0.000125

b.

the probability that atleast one channel is not playing a commercial

=1-the probability that at a given moment there are commercials playing at each channel

=1-0.000125

=0.999875

c.

lets visualize the problem,

for either channel 1 or channel 2 has a commercial playing at a given time the following situations come up.

for channel 1,

1. channel 1 playing, channel 2 not playing, channel 3 playing. P(X₁=1,X₂=0,X₃=1)
2. channel 1 playing, channel 2 not playing, channel 3 not playing. P(X₁=1,X₂=0,X₃=0)

for channel 2

1. channel 1 not playing, channel 2 playing, channel 3 playing. P(X₁=0,X₂=1,X₃=1)
2. channel 1 not playing, channel 2 playing, channel 3 not playing. P(X₁=0,X₂=1,X₃=0)

and we cannot take the cases when both channel 1 and channel 2 are playing

1. channel 1 playing, channel 2 playing, channel 3 playing. P(X₁=1,X₂=1,X₃=1)
2. channel 1 playing, channel 2 playing, channel 3 not playing. P(X₁=1,X₂=1,X₃=0)

so therefore,

P(either channel 1 or channel 2 has a commercial playing at a given time)

= P(X₁=1,X₂=0,X₃=1)P(X₁=1,X₂=1,X₃=0)+ P(X₁=0,X₂=1,X₃=1)P(X₁=0,X₂=1,X₃=0)-P(X₁=1,X₂=1,X₃=1) P(X₁=1,X₂=1,X₃=0)

=P(X₁=1)²P(X₂=0)²P(X₃=1)P(X₃=0)+P(X₁=0)²P(X₂=1)²P(X₃=1)P(X₃=0)-P(X₁=1)²P(X₂=1)²P(X₃=1)P(X₃=0)

=P(X₃=1)P(X₃=0)[P(X₁=1)²P(X₂=0)²+P(X₁=0)²P(X₂=1)²-P(X₁=1)²P(X₂=1)²]

=(1/20)(19/20)[(1/20²)(19/20)²+(19/20)²(1/20²)-(1/20²)(1/20²)]

=(19/400)[(1/400)(361+361-1)]

=(19/400)(721/400)

=13699/160000

=0.08561875

d.

the fourth channel is introduced which randomly gives 15 mins commercial in an hour,

let Y reprsents the fourth channel as,

P(Y=1)=15/60=1/4 and

P(Y=0)=45/60=3/4

the probability that at a given moment there are commercials playing at each channel (there is nothing to watch)

=P(X₁=1,X₂=1,X₃=1,Y=1)

=P(X₁=1).P(X₂=1).P(X₃=1).P(Y=1) [since independent]

=(1/20).(1/20).(1/20).(1/4)

=0.00003125