In a family of two children, find the probability that the family has two boys given that the first child is a boy.

A study determined that 5% of children under 18 years old lived with their father only. Find the probability that exactly 2 children selected at random from 12 children under 18 years old lived with their father only.

This is a binomial probability problem, where:

n = 12 (number of trials)
p = 0.05 (probability of success, which is living with father only)
x = 2 (number of successes)

To solve this problem, we can use the binomial probability formula:

P(x) = (n choose x) * p^x * (1-p)^(n-x)

where (n choose x) = n! / (x! * (n-x)!)

Plugging in the values:

P(2) = (12 choose 2) * 0.05^2 * 0.95^10
P(2) = (12! / (2! * 10!)) * 0.0025 * 0.5987
P(2) = 0.0117

Therefore, the probability that exactly 2 children selected at random from 12 children under 18 years old lived with their father only is 0.0117 or approximately 1.17%.

To find the probability that a family of two children has two boys given that the first child is a boy, we can use the concept of conditional probability.

Let's break down the possible outcomes:

1. Boy-Boy: BB
2. Boy-Girl: BG
3. Girl-Boy: GB
4. Girl-Girl: GG

We are given that the first child is a boy, which eliminates the last option (GG) as a possibility.

Out of the remaining three options, two have two boys (BB and BG). So the probability that the family has two boys given that the first child is a boy is 2 out of 3.

To calculate this as a probability, we divide the number of successful outcomes (2) by the total number of possible outcomes (3):

Probability = Number of Successful Outcomes / Total Number of Possible Outcomes

Therefore, the probability that the family has two boys given that the first child is a boy is 2/3 or approximately 0.67.