Respuesta :
Answer:
0.6 is the required probability.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 250
Probability of accessing internet at work = 37%
[tex]P(W)=0.37[/tex]
Probability of accessing internet at home = 44%
[tex]P(H)=0.44[/tex]
Probability of accessing internet at work and home = 21%
[tex]P(W\cap H)=0.21[/tex]
We have to evaluate the probability that a person in this sample accesses the internet at home or at work.
[tex]P(H\cup W) = P(H) + P(W) - P(W\cap H)\\P(W\cup H) = 0.37 + 0.44-0.21\\P(H\cup W) = 0.6[/tex]
Thus, 0.6 is the probability that a person in this sample accesses the internet at home or at work
Using Venn probabilities, it is found that there is a 0.6 = 60% probability that a person in this sample accesses the internet at home or at work.
The events are:
Event W: Person access internet at work.
Event H: Person access internet at home.
37% access the internet at work.
This means that [tex]P(W) = 0.37[/tex]
44% access the internet at home.
This means that [tex]P(H) = 0.44[/tex]
21% access the internet at both work and home.
This means that [tex]P(W \cap H) = 0.21[/tex]
What is the probability that a person in this sample accesses the internet at home or at work?
This probability is:
[tex]P(W \cup H) = P(W) + P(H) - P(W \cap H)[/tex]
Replacing the values:
[tex]P(W \cup H) = 0.37 + 0.44 - 0.21 = 0.6[/tex]
0.6 = 60% probability that a person in this sample accesses the internet at home or at work.
A similar problem is given at https://brainly.com/question/23508811
