An event occurring with absolute certainty.
The probability of having 2 boys and 1 girl in a family with three children can be calculated using the binomial probability formula. Assuming the probability of having a boy or a girl is equal (1/2 each), the probability of having 2 boys and 1 girl can be found by considering the different combinations (BBG, BGB, GBB). Therefore, the probability is ( \frac{3}{8} ) or 37.5%.
1:1
It is an absolute certainty.
Since the probability of having a son is about 1/2, the probability of the first 4 children being boys is about (1/2)4.
1-.015 = .985
Assuming that the probability of having a baby girl is 1/2 and that of having a baby boy is 1/2, the probability of having 3 baby girls in a row is (1/2)(1/2)(1/2)=1/8.
1:1
The probability of having 2 boys and 1 girl in a family with three children can be calculated using the binomial probability formula. Assuming the probability of having a boy or a girl is equal (1/2 each), the probability of having 2 boys and 1 girl can be found by considering the different combinations (BBG, BGB, GBB). Therefore, the probability is ( \frac{3}{8} ) or 37.5%.
1-.015 = .985
It is an absolute certainty.
9:3:3:1 The probability of having both recessive traits is 1:16.
Since the probability of having a son is about 1/2, the probability of the first 4 children being boys is about (1/2)4.
1-.015 = .985
The probability of having a girl versus a boy is 1/2 because there is two things you have a chance of getting and you can only get one or the other.
Assuming the probability of having a boy is 1/2, and that the probabilities are independent: Probability of 1 girl and 12 boys = (1/2)13 * 13 = 0.001587..., which is around 1/630
1/8
1/8?