bradenrio111
Answered

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According to a recent survey of adults, approximately 62% carry cash on a regular basis. The adults were also
asked if they have children. Of the 46% who have children, 85% carry cash on a regular basis. Is carrying cash
independent from having children in this sample?
O No, P(carry cash) = P(carry cash|have children).
O No, P(carry cash) + P(carry cash have children).
Yes, P(carry cash) = P(carry cash|have children).
Yes, P(carry cash) = P(carry cash have children).

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

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