Question

In Canada, 92% of the households have televisions.
72% of households have televisions and internet access.
5% have neither. Use appropriate probability notation to answer the following:

A) create a relative frequency table of this situation.

Answer 1

Answer: Let "T" denote households with television

"I" denote households with internet

and "N" denote households with neither of these.

Since P(N) = 0.05 then it means P(TuI) = 0.95. We know from set theory that P(TuI) = P(T) + P(I) - P(TnI). Thus,

0.95 = 0.92 + 0.72 - P(TnI) ==> P(TnI) = 0.69. This means P(T/I) = P(T) - P(TnI) = 0.92 - 0.69 = 0.23, and P(I/T) = P(I) - P(TnI) = 0.72 - 0.69 = 0.03. So,

a)

House                 Probability

Nothing                0.05

Only television      0.23

Only internet         0.03

Both                   0.69

b) P(internet but no television) = P(only internet) = 0.03

c) P(internet | there is television) = (By Bayes' Rule) P(TnI) / P(T) = 0.69 / 0.95 = 0.726 (rounded to 3 decimal points)

d) P(internet | there is no television) = (By Bayes' Rule) P(I/T) / P(T') = 0.03 / 0.08 = 0.375

e) From (c) and (d) ==> In Canada, it is (0.726/0.375 ≈ 2 ) twice more likely that if there's television in the household, then there is internet.

Step-by-step explanation: