Answer to Question #202337 in Statistics and Probability for Aasia Tariq

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 two or more TV sets?

d. Why does the special law of addition not apply to this problem?

Answer 1

Solution:

Notations:

C - televisions with cable TV

M - televisions with 2 or more TV sets

Given, $P(C)=67\%,P(M)=74\%,P(C\cap M)=55\%$

(a) $P(C\cup M)=P(C)+P(M)-P(C\cap M)$

$=(67+74-55)\% =86\%$

(b) $P(C\cup M)-P(C \cap M)=86\%-55\%=31\%$

(c) $P(C'\cap M')=1-P(C \cup M)=100\%-86\%=14\%$

(d) Special law of addition is: $P(A\cup B)=P(A)+P(B)$

Here, A and B are mutually exclusive events.

It is not applied here as the given two events are not mutually exclusive events.