This is a binomial probability distribution with the following criteria: Number of trials = 6, number of successes = 2, probability of success = 1/6 or 0.16667. Because of the probability value not being a typical number, there is not a table to go to look the answer up. The math is tedious. So, we go to a binomial probability calculator and enter in the aforementioned data to obtain the answer of 0.201.
2/4
There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.
0.1%
50%
0.25 binomial distribution, where p=0.5, q=0.5, x=3, n=4 4!/(3!*1!)*0.530.51 = 0.25 Also can be solved by identifying each event possible and related probability. There are 4 ways this can occur (first child is a girl, second child is a girl, third child is a girl and fourth child is a girl) and there is a 0.54 chance of each of these events occurring. Prob= 4 *0.0625 = 0.25
The probability is 2 - 6
2/4
Start with examples like flipping a coin, rolling a die or spinning a dreidel. Then explain in terms they understand. That depends very much on the age of the child.
This is a Binomial Probability; p=0.5, n=10 & x=7. Since you want the probability of exactly 7, in the related link calculator, after placing in the above values, P(x=7) = 0.1172 or 11.72%.
No probability. Neither parent has an "A" for the child to inherit to make an "AB".
There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.There is not enough information on the propensity for the parents to have a child of either gender and so it is necessary to assume that the probability of the gender of the next child is independent of the genders of preceding children. In that case the probability of the next child being a girl is 1/2.
1 in 2
It depends on the context: if you select a child at random from a girls' school, the probability is 0, while if it is at a boys' school it is 1!
The probability is zero! There is no such thing as "normal". Every child (and adult) has some unique characteristics and that makes them not normal - in that respect.
Since having a child to a child is an independent event (assuming no outside intervention), the probability is still about 50 / 50 boy or girl.
The child will have the disorder, only if the recessive allele from both the parents is transferred to the child. Therefore, the probability is 1/4.
If the gender of a child were an independent variable then the genders of the existing children would be irrelevant and so the probability of the next child being a girl would be approximately 1/2.It would be approximately 1/2 because the overall proportion is not exactly half. However, and more important, is the fact that the gender of a child is affected by the parents' genes and so is not independent of the gender of previous children.