P(subscribes to at least one of the two services) = P(A U B) = P(A) + P(B) – P(A N B) = .6 + .8 – 5 = 9 The event that a household subscribes only to tv service can be written as A' N B [(not Internet) and TV]. Now Figure 2.4 implies that .9 = P(A U B) = P(A) + P(A' N B) = .6 + P(A' N B) from which P(A' n B) = .3. Similarly, P(A N B') = P(A U B) – P(B) = .1. This is all illustrated in Figure 2.5, from which we see that P(exactly one) = P(A N B') + P(A' N B) = .1 + .3 = 4 P(A N B') P(A'n B) .3
Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
In a certain residential suburb, 60% of all households get Internet service from the
local cable company, 80% get television service from that company, and 50% get
both services from that company. If a household is randomly selected, what is the
probability that it gets at least one of these two services from the company, and what
is the probability that it gets exactly one of these services from the company?
With A 5 {gets Internet service} and B 5 {gets TV service}, the given information
implies that P(A) 5 .6, P(B) 5 .8, and P(A ù B) 5 .5. The foregoing proposition
now yields
![P(subscribes to at least one of the two services)
= P(A U B) = P(A) + P(B) – P(A N B) = .6 + .8 – 5 = 9
The event that a household subscribes only to tv service can be written as A' N B
[(not Internet) and TV]. Now Figure 2.4 implies that
.9 = P(A U B) = P(A) + P(A' N B) = .6 + P(A' N B)
from which P(A' n B) = .3. Similarly, P(A N B') = P(A U B) – P(B) = .1. This is
all illustrated in Figure 2.5, from which we see that
P(exactly one) = P(A N B') + P(A' N B) = .1 + .3 = 4
P(A N B')
P(A'n B)
.3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff22117d4-84e4-44cb-be6c-aa70fe2734d4%2F388b7b10-180e-402f-8680-7e032f029393%2F3trcyb.png&w=3840&q=75)
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