Question

3. fifty - eight percent of sri lankan children (ages 3-5) are trained to read to everyday by family members. suppose 5 children are randomly selected. what is the probability that at least one is trained everyday by family members at home?

Answer 1

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.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

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:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.58)^{0}.(0.42)^{5} = 0.0131$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0131 = 0.9869$

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 brainly.com/question/24863377

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