Question

The Wilson family was one of the first to come to the U.S. They had 8 children. Assuming that the

Answer 1

Answer:

0.3634 = 36.34% probability that the Wilson family had at least 5 girls.

0.8554 = 85.54% probability that the Wilson family had at most 5 girls.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it is a girl, or it is not. The probability of a child being a girl is independent of any other child. This means that the binomial probability distribution is used 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 8 children.

This means that $n = 8$

The probability of a child being a girl is .5

This means that $p = 0.5$

Probability that the Wilson family had: at least 5 girls?

This is:

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)$

In which

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

$P(X = 5) = C_{8,5}.(0.5)^{5}.(0.5)^{3} = 0.2188$

$P(X = 6) = C_{8,6}.(0.5)^{6}.(0.5)^{2} = 0.1094$

$P(X = 7) = C_{8,7}.(0.5)^{7}.(0.5)^{1} = 0.0313$

$P(X = 8) = C_{8,8}.(0.5)^{8}.(0.5)^{0} = 0.0039$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2188 + 0.1094 + 0.0313 + 0.0039 = 0.3634$

0.3634 = 36.34% probability that the Wilson family had at least 5 girls.

At most 5 girls?

This is:

$P(X \leq 5) = 1 - P(X > 5)$

In which

$P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) = 0.1094 + 0.0313 + 0.0039 = 0.1446$

$P(X \leq 5) = 1 - P(X > 5) = 1 - 0.1446 = 0.8554$

0.8554 = 85.54% probability that the Wilson family had at most 5 girls.