Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
According to the probabilities given, it is found that the correct option regarding the independence of the events is given by:
No, P(carry cash) != P(carry cash|have children).
What is the probability of independent events?
If two events, A and B, are independent, we have that:
[tex]P(A \cap B) = P(A)P(B)[/tex]
Which also means that:
[tex]P(A|B) = P(A)[/tex]
[tex]P(B|A) = P(B)[/tex]
In this problem, we have that:
- 62% carry cash on a regular basis, hence P(cash) = 0.62.
- 46% has children, hence P(children) = 0.46.
- Of the 46% who have children, 85% carry cash on a regular basis, hence P(cash|children) = 0.85.
Since P(carry cash) != P(carry cash|have children), they are not independent.
More can be learned about the probability of independent events at https://brainly.com/question/25715148
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.