Question

The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Blair family had... (a) ...at least 3 girls

Answer 1

Answer:

31.25% probability that the Blair family had at least 3 girls

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

They had 4 children.

This means that $n = 4$

The probability of a child being a girl is 0.5

This means that $p = 0.5$

Probability of at least 3 children:

$P(X \geq 3) = P(X = 3) + P(X = 4)$

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

$P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25$

$P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125$

31.25% probability that the Blair family had at least 3 girls