Answer to Question #202894 in Statistics and Probability for Mehran khan

Question

According to Nielsen Media Research, approximately 67% of all U.S. households with television have cable TV. Seventy-four percent of all U.S. households with television have two or more TV sets. Suppose 55% of all U.S. households with television have cable TV and two or more TV sets. A U.S. household with television is randomly selected.

a. What is the probability that the household has cable TV or two or more TV sets?

b. What is the probability that the household has cable TV or two or more TV sets but not both?

c. What is the probability that the household has neither cable TV nor

Answer 1

A=households with television have cable TV

P(A) = 0.67

B=households with television have two or more TV sets

$P(B) = 0.74 \\ P(A \cap B) = 0.55$

a.$P(A \cup B) = P(A) + P(B) -P(A \cap B)$

=0.67 +0.74 - 0.55

=0.86

b. P(A or B but not both) $= P(A \cup B) - P(A \cap B)$

= 0.86 -0.55

= 0.31

c. P(neither A nor B) $= [P(A \cup B)^c]$

$= 1 - P(A \cup B)$

= 1 - 0.86

= 0.14