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 each of their offspring will develop the disease is approximately 0.25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on four 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? (Round your answers to six decimal places.) (a) all four children will develop Tay-Sachs disease. (b) only one child will develop Tay-Sachs disease. (c) the fourth child will develop Tay-Sachs disease, given that the first three did not.

Answer 1

Answer:

Step-by-step explanation:

Given that 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 each of their offspring will develop the disease is approximately 0.25.

Let X be the no of children having this disease out of 4 occasions

Then X has two outcomes and each trial is independent of the other trial

Hence X is binomial with n =4, and p = 0.25

a) $P(X=4) = 4C4 (0.25)^4 = 0.003906$

b) $P(X=1) = 4C1 (0.25)(0.75)^3\\= 0.4219$

c) $P(4th child/3 children did not) = 0.25$ since independent.