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In a family of eight children, what is the probability that (a) the third child is a girl? (b) six of the children are boys? (c) all the children are girls? (d) there are four boys and four girls? Assume that the probability of having a boy is equal to the probability of having a girl \((p=1 / 2)\).

Short Answer

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Question: Calculate the probabilities for each of the following scenarios in a family with eight children, with equal probability of having a boy or a girl: (a) The third child is a girl (b) Six of the children are boys (c) All the children are girls (d) There are four boys and four girls in the family Answer: (a) 1/2 (b) \(\binom{8}{6} (1/2)^8\) (c) \(\binom{8}{8} (1/2)^8\) (d) \(\binom{8}{4} (1/2)^8\)

Step by step solution

01

Calculate the probability that six of the children are boys

Using the binomial probability formula with n=8, k=6, and p=1/2, we get: \(P(X=6) = \binom{8}{6} (1/2)^6 (1/2)^{8-6} = \binom{8}{6} (1/2)^8\) Calculate this value to find the probability. (c) What is the probability that all the children are girls? In this scenario, we want to find out the probability that all 8 children are girls. We will use the binomial probability formula with n=8, k=8, and p=1/2.
02

Calculate the probability that all the children are girls

Using the binomial probability formula, we get: \(P(X=8) = \binom{8}{8} (1/2)^8 (1/2)^{8-8} = \binom{8}{8} (1/2)^8\) Calculate this value to find the probability. (d) What is the probability that there are four boys and four girls in the family? In this scenario, we want to find out the probability of having exactly 4 boys and 4 girls in the family. We will use the binomial probability formula with n=8, k=4, and p=1/2.
03

Calculate the probability of having four boys and four girls

Using the binomial probability formula, we get: \(P(X=4) = \binom{8}{4} (1/2)^4 (1/2)^{8-4} = \binom{8}{4} (1/2)^8\) Calculate this value to find the probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Having a Boy or Girl
Understanding the probability of having a boy or girl is essential when discussing genetics and family planning. In genetics, the sex of a child is determined by the X and Y chromosomes, with XX resulting in a girl and XY in a boy. Simplified assumptions consider that the chance of having either a girl or a boy is an independent event with a probability of 0.5 or 50%.

When we look at a specific child in a family, irrespective of the gender of any other children, the probability of that child being a girl is always 0.5. This constant probability holds true for each child due to the random combination of the parents' chromosomes.
Binomial Distribution
The binomial distribution is a statistical tool used to represent the number of successes in a fixed number of independent trials, with only two possible outcomes, usually termed success and failure. This distribution is defined by two parameters: the number of trials (n) and the probability of success (p) in each trial.

In the context of family planning, where 'success' might denote the birth of a boy (or a girl, depending on the question), the binomial distribution provides a formula to calculate the probability of having a certain number of boys (or girls) in a specified number of children. The formula for the binomial probability is: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(P(X=k)\) is the probability of k successes (e.g., k boys) in n trials (births), and \(p\) is the probability of a success in a single trial.
Genetic Probability Calculations
To perform genetic probability calculations, one must understand not only simple occurrence probabilities (such as the 50% chance of having a boy) but also how these probabilities interact across multiple events. The rules of probability, including multiplication and addition principles, play a critical role in these calculations.

The multiplication rule is used for finding the probability of the intersection of two independent events, like having a sequence of children of particular genders. For instance, the chance of having three boys consecutively is calculated by multiplying the probability of having a boy three times: \(0.5 \times 0.5 \times 0.5\). This is essential to understanding more complex genetic inheritance patterns and can be extended to a range of genetic events.
Combinatorics in Genetics
Combinatorics, the branch of mathematics concerning counting, is also instrumental when dealing with genetics, especially for calculating probabilities in larger family scenarios. Combinatorial analysis helps to define the number of ways events can occur and is crucial for understanding the binomial distribution.

For example, the binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose k successes out of n trials, which is why it appears in the binomial probability formula. So, when calculating the chance of having four boys and four girls in an eight-child family, we use combinatorics to determine that there are \( \binom{8}{4} \) ways to arrange those genders. This figure is then incorporated into the formula to calculate the overall probability.

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Most popular questions from this chapter

In assessing data that fell into two phenotypic classes, a geneticist observed values of \(250: 150 .\) She decided to perform a \(\chi^{2}\) analysis by using the following two different null hypotheses: (a) the data fit a 3: 1 ratio, and (b) the data fit a 1: 1 ratio. Calculate the \(\chi^{2}\) values for each hypothesis. What can be concluded about each hypothesis?

Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).

The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the \(p\) value used to reject or not reject the hypothesis. In the case of the chi- square analysis of genetic crosses, would the use of a standard of \(p=0.10\) be more or less stringent about not rejecting the null hypothesis? Explain.

Dentinogenesis imperfecta is a tooth disorder involving the production of dentin sialophosphoprotein, a bone-like component of the protective middle layer of teeth. The trait is inherited as an autosomal dominant allele located on chromosome 4 in humans and occurs in about 1 in 6000 to 8000 people. Assume that a man with dentinogenesis imperfecta, whose father had the disease but whose mother had normal teeth, married a woman with normal teeth. They have six children. What is the probability that their first child will be a male with dentinogenesis imperfecta? What is the probability that three of their six chil- dren will have the disease?

Tay-Sachs disease (TSD) is an inborn error of metabolism that results in death, often by the age of \(2 .\) You are a genetic counselor interviewing a phenotypically normal couple who tell you the male had a female first cousin (on his father's side) who died from TSD and the female had a maternal uncle with TSD. There are no other known cases in either of the families, and none of the matings have been between related individuals. Assume that this trait is very rare. (a) Draw a pedigree of the families of this couple, showing the relevant individuals. (b) Calculate the probability that both the male and female are carriers for TSD. (c) What is the probability that neither of them is a carrier? (d) What is the probability that one of them is a carrier and the other is not? [Hint: The \(p\) values in (b), (c), and (d) should equal \(1 .]\)

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