Suppose that a certain city has three major newspapers the T

Suppose that a certain city has three major newspapers, the Times, Herald, and Examiner. Circulation indicates 47.0% of households get the Times, 33.4% get the Herald, 34.6% get the Exam- iner, 11.9% get the Times and Herald, 15.1% get the Times and Examiner, 10.4% get the Herald and Examiner and 4.8% get all three. If a household is chosen at random determine the probability that it gets none of the three major newspaper?

Solution

Let T shows the event that household get Times, H shows Herald and E shows Examiner so we have follwoing probabilities:

P(T)=0.47

P(H)=0.334

P(E)=0.346

P(T and H)=0.119

P(T and E)=0.151

P(H and E)=0.104

P(T and H and E)=0.048

Let us find the probability that household get atleast one of the three major newspaper:

P(T or H or E)= P(T)+P(H)+P(E)-P(T and H)- P(H and E)- P(T and E)+P(T and H and E)

=0.47+0.334+0.346-0.119-0.151-0.104+0.048=0.824

So the probability that it gets none of the three major newspaper will be

1- P(T or H or E)=1-0.824=0.176

Hence, the required probability is 0.176.