Question

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, given that the first two did not.

Short answer

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, given that the first two did not. Answer: a. 0.015625 b. 0.421875 c. 0.25

Step-by-step solution

Define Constants

Let P(T) be the probability of a child developing Tay-Sachs disease. Given P(T) = 0.25. Let P(NT) be the probability of a child not developing Tay-Sachs disease. Then, P(NT) = 1 - P(T) = 1 - 0.25 = 0.75 For each scenario, we will calculate the probability using the given information: #a. All three children develop Tay-Sachs disease:

Calculate the Probability of All Three Developing Tay-Sachs

Since the occurrences in each offspring are independent, the probability of all three children developing the disease is the product of the probabilities of each child developing the disease: P(T₁ ⋂ T₂ ⋂ T₃) = P(T) × P(T) × P(T) = 0.25 × 0.25 × 0.25 = 0.015625 #b. Only one child will develop Tay-Sachs disease:

Calculate the Probability of Only One Child Developing Tay-Sachs

There are three possible cases for only one child developing Tay-Sachs: the first child, the second child, or the third child. Since each case is mutually exclusive, we add their probabilities: P(Only one child has Tay-Sachs) = P(T₁ ⋂ NT₂ ⋂ NT₃) + P(NT₁ ⋂ T₂ ⋂ NT₃) + P(NT₁ ⋂ NT₂ ⋂ T₃) = P(T) × P(NT) × P(NT) + P(NT) × P(T) × P(NT) + P(NT) × P(NT) × P(T) = 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. The third child will develop Tay-Sachs disease, given that the first two did not:

Calculate the Conditional Probability of the Third Child Developing Tay-Sachs

We need to find the conditional probability of the third child developing Tay-Sachs given that the first two did not. Since the occurrences in each offspring are independent, this conditional probability is just the probability of the third child developing the disease regardless of the first two children's condition, which is: P(T₃ | NT₁ ⋂ NT₂) = P(T) = 0.25 The probabilities for each scenario are: a. 0.015625 b. 0.421875 c. 0.25

Key concepts

Genetic Disorders

Genetic disorders are health conditions caused by abnormalities in an individual's DNA. These disorders can be inherited from parents if one or both carry changed genes related to a specific condition.
Tay-Sachs disease is a perfect example of an autosomal recessive genetic disorder. This means both parents must carry one copy of the mutated gene for their child to develop the disease. If both parents are carriers, there is a 25% chance with each pregnancy that their child will inherit the disorder.
It's important to note that being a carrier means having one normal and one mutated gene copy. Carriers themselves do not usually show symptoms, but they can pass the mutated gene on to their offspring. In this context, genetic counseling can be beneficial for families who are known carriers and are concerned about passing on the disorder to their children. This helps them understand and manage the risks associated with the disorder.

Independent Events

Understanding independent events can simplify the calculation of probabilities in complex scenarios. In probability theory, two events are considered independent if the occurrence of one does not affect the occurrence of the other.
In the case of Tay-Sachs disease, the chances of each child developing the disease are independent of one another, meaning whether one child has the disease does not influence the probability of another child having it. This simplifies calculations because you can use the same probability for each child, independently multiplying their individual probabilities for events such as all children contracting the disorder.
For example, the probability of all three children developing Tay-Sachs disease is computed by multiplying the probability for each child: \[ P(T₁ \text{ and } T₂ \text{ and } T₃) = 0.25 \times 0.25 \times 0.25 = 0.015625 \] This demonstrates how key independent events are for straightforward multiplication of probabilities.

Conditional Probability

Conditional probability helps us find the probability of an event occurring given that another event has already occurred. It's especially useful when dealing with dependent events but can also clarify independent event scenarios.
In our example, we calculate the probability that the third child develops Tay-Sachs disease (event T₃), given that the first two did not (event NT₁ ∩ NT₂). Independent events simplify this calculation. Despite the first two children not developing the disease, the event does not influence the probability of the third child being affected, which remains at 25%. Thus, \[P(T₃ | \text{NT₁ ∩ NT₂}) = P(T) = 0.25 \]This calculation shows how independence in probability allows the condition set by the first two outcomes to not affect the third child's probability of having Tay-Sachs, as long as the events remain independent.