Using the binomial distribution, it is found that there is a 0.9869 = 98.69% probability that at least one is trained everyday by family members at home.
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem, the values of the parameters are given as follows:
n = 5, p = 0.58.
The probability that at least one is trained everyday by family members at home is given by:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.58)^{0}.(0.42)^{5} = 0.0131[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0131 = 0.9869[/tex]
0.9869 = 98.69% probability that at least one is trained everyday by family members at home.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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