Question

Assume each newborn baby has a probability of approximately 0.55 of being female and 0.45 of being male. for a family with three ​children, let x equal number of children who are girls.

a. identify the three conditions that must be satisfied for x to have the binomial distribution.
b. identify n and p for the binomial distribution.
c. find the probability that the family has one girl and two boys.

Answer 1

1. There must be exactly two possible outcomes (girl, boy).

2. Repeated trials must be independent.

3. The probabilities of the two outcomes don't change from trial to trial.

n = 3 (there are 3 children "generated")

p = 0.55 (the probability that a child is a girl)

The probability that a family of 3 consists of one girl and two boys is

`_3C_1 * (.55)^1 * (.45)^2 = 3 * .55 * .2025 = 0.334125`

In general, the probability that a binomially distributed random variable X is equal to r is

`_nC_r * p^r * (1-p)^(n-r)`