Question

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.

Answer 1

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

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

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

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:

• 3 children, thus $n = 3$
• Equally as likely to be a girl or a boy, thus $p = 0.5$.

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

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

$P(X = 1) = C_{3,1}.(0.5)^{1}.(0.5)^{2} = 0.375$

0.375 = 37.5% probability of having exactly 1 girl.

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