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zubka84 [21]
2 years ago
6

Determine the probability of selecting a two-child family with one boy and one girl assuming boys and girls are equally likely

Mathematics
1 answer:
Fittoniya [83]2 years ago
3 0

Using the binomial distribution, it is found that there is a 0.5 = 50% probability of selecting a two-child family with one boy and one girl.

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

Binomial probability distribution

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:

  • Two children, hence n = 2.
  • Equally as likely to be a boy or a girl, hence p = 0.5.

The probability of one of each is P(X = 1), hence:

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

P(X = 1) = C_{2,1}.(0.5)^{1}.(0.5)^{1} = 0.5

0.5 = 50% probability of selecting a two-child family with one boy and one girl.

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

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\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 }  =  \frac{ { \sin}^{2} y}{ { \tan }^{2} y}  =  \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } }  =  { \cos }^{2} y

6 0
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I turned it in and it says we got one wrong(out of those blanks). I've checked over it several times but I can't figure out whic
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Answer:

Try as percentages so

7) 37.5%

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9) 12.5%

10) 62.5%

11) 87.5%

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Consider the missing lines of question are "Due to the Enhanced Community Quarantine, Grand Royal Spa temporarily stopped its operation and to help the employees, the owner decided to split evenly its total revenue of 65,000.00".

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The function which represents the  amount each received.

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