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in a family with 3 children, excluding multiple births, what is the probability of having exactly 1 girl? assume that having a boy is as likely as having a girl at each birth.

Sagot :

Using the binomial distribution, it is found that there is a 0.375 = 37.5% probability of having exactly 1 girl.

For each children, there are only two possible outcomes. Either it is a boy, or it is a girl. The probability of a children being a girl is independent of any other children, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[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:

  • 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:

  • 3 children, thus [tex]n = 3[/tex]
  • Equally as likely to be a girl or a boy, thus [tex]p = 0.5[/tex].

The probability of exactly 1 girl is P(X = 1), thus:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{3,1}.(0.5)^{1}.(0.5)^{2} = 0.375[/tex]

0.375 = 37.5% probability of having exactly 1 girl.

A similar problem is given at https://brainly.com/question/24863377

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