Question

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 Examiner, 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 at least one of the three major newspaper?

Answer 1

0.824

Explanation:

Given:

Household get Times P(T) = 0.47

Household get Herald, P(H) = 0.334

Household get Examiner, P(E) = 0.346

Household get the Times and Herald, P(T and H)=0.119

Household get the Times and Examiner, P(T and E)=0.151

Household get the Herald and Examiner, P(H and E)=0.104

Household get all three, P(T and H and E)=0.048

Now,

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