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

Answer: The probabilities for the three scenarios are: a) 0.015625, b) 0.421875, and c) 0.25.

Step-by-step solution

a. All three children develop Tay-Sachs disease

Since all three children developing the disease are independent events, we can multiply the probability of one child developing the disease by itself three times, or simply cube the probability. This calculation is as follows: (0.25)^3 = 0.015625 So, the probability of all three children developing Tay-Sachs disease is 0.015625.

b. Only one child develops Tay-Sachs disease

For this scenario, we need to consider the three possibilities of only one child developing the disease: either the first, second, or third child. We can compute the probability for each of these situations and then sum them up. Let's compute the probability for each child: 1st child: (0.25) * (1 - 0.25)^2 = 0.25 * 0.5625 = 0.140625 2nd child: (1 - 0.25) * (0.25) * (1 - 0.25) = 0.140625 3rd child: (1 - 0.25)^2 * (0.25) = 0.140625 Now sum the probabilities: 0.140625 + 0.140625 + 0.140625 = 0.421875 So, the probability of only one child developing Tay-Sachs disease is 0.421875.

c. The third child develops Tay-Sachs disease, given that the first two did not

To find this conditional probability, we can apply the basic probability rules. We know that the probability of the third child developing the disease is 0.25. Since the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, the fact that the first two did not develop the disease has no impact on the probability for the third child. Therefore, the probability remains the same: P(third child develops Tay-Sachs | first two did not) = 0.25

Key concepts

Tay-Sachs Disease Inheritance

Tay-Sachs disease is a tragic genetic disorder primarily affecting the nerve cells in the brain and spinal cord. It is inherited in an autosomal recessive manner, which means that a person must receive one defective gene from each parent to be affected by the disease.

When parents are carriers of the Tay-Sachs gene, they do not show symptoms themselves because they also possess a normal copy of the gene. Each child of carrier parents has a 25% chance of inheriting two defective genes (one from each parent), which would result in the development of the condition; a 50% chance of being a carrier like their parents, with one normal and one defective gene; and a 25% chance of inheriting two normal genes. This genetic probability remains constant for each child, regardless of the genetic outcome in their siblings. The misconception that the risk decreases with each subsequent child is a common confusion and should be avoided for accurate understanding.

Independent Events Probability

Independence in probability refers to events whose outcomes do not influence each other. When calculating the probability of multiple independent events occurring, we multiply their individual probabilities together. This is key to understanding scenarios such as the chances of multiple children inheriting Tay-Sachs when each child's genetic outcome is independent from their siblings'.

For example, when we say that the probability of all three children developing Tay-Sachs disease is \( (0.25)^3 = 0.015625 \), we assume the outcome of one child does not affect the others. This assumption of independence is crucial for computations in genetics, as well as in many other areas where multiple stochastic events occur.

Conditional Probability

Conditional probability assesses the likelihood of an event given that another event has already occurred. In the context of genetics and Tay-Sachs disease, it can describe scenarios such as the probability of a child having the disease given that their siblings do not.

In the provided exercise, the conditional probability does not differ from the original probability of the third child developing Tay-Sachs disease, which is 0.25. This is because the independence of each child's genetic outcome means the previous births do not change the probability for the third child. Understanding this distinction between dependent and independent events is essential for applying the correct methods in probability calculations.