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Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately .25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? a. All three children will develop Tay-Sachs disease. b. Only one child will develop Tay-Sachs disease. c. The third child will develop Tay-Sachs disease, siven that the first two did

Short Answer

Expert verified
Answer: 0.25

Step by step solution

01

Identify the probability of each child developing Tay-Sachs disease

The exercise mentions that the probability an offspring will develop Tay-Sachs disease if both parents are carriers is 0.25. Therefore, the probability for each child is 0.25.
02

Multiply the probabilities for independent events

Since the occurrence of Tay-Sachs in any offspring is independent, we can multiply the probabilities for each child to obtain the result. For three children developing the disease: P(A and B and C) = P(A) * P(B) * P(C) = (0.25) * (0.25) * (0.25) = 0.015625. b. To find the probability that only one child will develop Tay-Sachs disease:
03

Identify the probability of success and failure

Like mentioned previously, given both parents are carriers, the probability a child will develop Tay-Sachs is 0.25. Consequently, the probability a child will not develop the disease is 0.75 (1 - 0.25).
04

Use combinations to account for different scenarios

There are 3 possible scenarios: (1) Only the first child develops the disease, (2) only the second child develops the disease, (3) only the third child develops the disease. For each of these scenarios, multiply the probabilities of success and failure. (1) First child: (0.25)(0.75)(0.75) (2) Second child: (0.75)(0.25)(0.75) (3) Third child: (0.75)(0.75)(0.25)
05

Add up the probabilities for each scenario

Sum the probabilities for all three scenarios: (0.25)(0.75)(0.75) + (0.75)(0.25)(0.75) + (0.75)(0.75)(0.25) = 3 * (0.25)(0.75)(0.75) = 0.421875. c. To find the probability that the third child will develop Tay-Sachs disease, given that the first two did:
06

Understand the conditional probability

We want to find the probability of the third child developing Tay-Sachs disease, given that the first two already have the disease. Since the occurrences are independent, the probability of the third child developing the disease remains unchanged, irrespective of the first two children.
07

Identify the probability of the third child developing Tay-Sachs disease

The probability of the third child developing Tay-Sachs (given both parents are carriers) is 0.25, and it is not affected by the condition of the first two children. Therefore, the probability of the third child developing the Tay-Sachs disease given the first two did is still 0.25.

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Key Concepts

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

Tay-Sachs Disease Inheritance
Tay-Sachs disease is an autosomal recessive genetic condition, which means that a child must inherit a defective gene from each parent to exhibit symptoms of the disease. The mechanics of inheritance are critical to understanding the probabilities involved in genetic disorders like Tay-Sachs.

In cases where both parents are carriers of the Tay-Sachs disease gene, each of their children has a probability of inheriting the condition. This is calculated based on Mendelian genetics, where each parent contributes one of two possible alleles for a gene. In simple terms, the possible combinations are:
  • Both defective genes are inherited (disease develops),
  • One defective and one healthy gene (carrier without symptoms),
  • Both genes are healthy (no disease and not a carrier).
When both parents are carriers, the probabilistic distribution for their child is 25% to inherit and exhibit Tay-Sachs, 50% to be a carrier, and 25% to be completely unaffected.

The key point to remember is that each pregnancy is an independent event with the same probability distribution, assuming the genetic makeup of the parents remains constant. This understanding lays the foundation for analyzing specific probability scenarios.
Independent Events in Probability
An understanding of independent events is foundational to calculating probabilities in genetics and many other fields. Essentially, two events are independent if the occurrence of one event does not impact the probability of the other occurring. In the case of Tay-Sachs disease, the genetic makeup of one child does not influence the genetic combinations of the next child.

Therefore, when calculating the probability of different birth outcomes for multiple children of the same parents, the rules of independent events let us multiply the probability of each event occurring separately. For example, if the probability of an event A is p(A) and event B is p(B), and A and B are independent, then the probability of both A and B occurring, denoted as p(A and B), is p(A) * p(B).

It is essential when tackling problems to recognize that the independence of events simplifies the process of finding combined probabilities. This can be particularly empowering for students to understand complex genetic probability questions.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already taken place. This concept plays a massive role in genetic probability, as it can help us predict future outcomes based on previous ones, even though, in the case of Tay-Sachs inheritance, the events are independent.

To determine a conditional probability, we typically divide the probability of both events happening by the probability of the condition being met. However, when events are independent, as in the birth of children with Tay-Sachs disease with carrier parents, the probability of an event is not influenced by prior outcomes.

Therefore, the concept of conditional probability adapts slightly. As seen in the exercise, the third child's chances of developing Tay-Sachs remain consistent (25%) regardless of the first two children's health outcomes. Understanding conditional probability is a building block for more complex probability problems and is an essential tool for students learning genetics or other disciplines involving uncertainty.

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Most popular questions from this chapter

Most weather forecasters protect themselves very well by attaching probabilities to their forecasts, such as "The probability of rain today is \(40 \%\)." Then, if a particular forecast is incorrect, you are expected to attribute the error to the random behavior of the weather rather than to the inaccuracy of the forecaster. To check the accuracy of a particular forecaster, records were checked only for those days when the forecaster predicted rain "with \(30 \%\) probability." A check of 25 of those days indicated that it rained on 10 of the \(25 .\) a. If the forecaster is accurate, what is the appropriate value of \(p,\) the probability of rain on one of the 25 days? b. What are the mean and standard deviation of \(x,\) the number of days on which it rained, assuming that the forecaster is accurate? c. Calculate the \(z\) -score for the observed value, \(x=10 .\) [HINT: Recall from Section 2.6 that \(z\) -score \(=(x-\mu) / \sigma .\) d. Do these data disagree with the forecast of a "30\% probability of rain"? Explain.

Find the mean and standard deviation for a binomial distribution with these values: a. \(n=1000, p=.3\) b. \(n=400, p=.01\) c. \(n=500, p=.5\) d. \(n=1600, p=.8\)

Use the formula for the binomial probability distribution to calculate the values of \(p(x)\), and construct the probability histogram for \(x\) when \(n=6\) and \(p=.2\). [HINT: Calculate \(P(x=k)\) for seven different values of \(k\).]

A candy dish contains five blue and three red candies. A child reaches up and selects three candies without looking. a. What is the probability that there are two blue and one red candies in the selection? b. What is the probability that the candies are all red? c. What is the probability that the candies are all blue?

Records show that \(30 \%\) of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose \(n=4\) new patients represent a random selection from the large set of prospective patients served by the clinic. Find these probabilities: a. All the patients' bills will eventually have to be forgiven. b. One will have to be forgiven. c. None will have to be forgiven.

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