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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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