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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).
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:
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