Question

Tay–Sachs Disease 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 ofthese 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.

Answer 1

Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that $p = 0.25$

Three children:

This means that $n = 3$

Question a:

This is P(X = 3). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156$

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219$

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

$p = 0.75*0.75*0.25 = 0.1406$

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.