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