Question

In Happyville, the ideal family has four children, at least two of which are girls. If you randomly select a Happyville family that has four children, what is the probability that it is not ideal

Answer 1

Answer:

0.3125 = 31.25% probability that it is not ideal.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it is a girl, or they are 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.

0.5 probability of a children being a girl:

This means that $p = 0.5$

If you randomly select a Happyville family that has four children, what is the probability that it is not ideal?

Not ideal is less than two girls, so this is P(X < 2).

Four children means that $n = 4$. So

$P(X < 2) = P(X = 0) + P(X = 1)$

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

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

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

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0625 + 0.25 = 0.3125$

0.3125 = 31.25% probability that it is not ideal.