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

INCOMPLETE QUESTION:

The events we have to calculate probability are

a) All childrean will develop Tay-Sachs

b) Only one child will develop Tay-Sachs

c)The third child will develop Tay-Sachs, given that the first two did not

a) The probability that the dissease is developed by all four children is 1/256

b) The probability that exactly one child develops Tay-Sachs is 27/64

c) The probability that the third child will develop Taay-Sachs is 1/4, given that the first two child did develop it.

Step-by-step explanation:

a) Each child can develop the dissease with probability 0.25. Since each child's outcome is independent on the other, we can calculate the probability that the 4 children get the illnes by powering 0.25 py 4, thus

P(All children develop Tay-Sachs) = 0.25⁴ = 1/256

b) There are 4 ways to select the child who will develop the dissease, and for each way, the child has a probability of 0.25 of deveoping it while the other 3 has a probability of 0.75 of not doing so. Since the events are independent on each other, the probability that exactly one child develops the dissease is

4 * (0.25)*(0.75)³ = 27/64

c) Since the occurrence of Tay-Sachs in each child in independent from the others, then the probability that the third child develops Tay-Sachs isnt modified if we know that the first two offspring got the dissease. Therefore, the probability that the third child will develop Taay-Sachs is 1/4 = 0.25.

60%

answer is because 4×10=40

and 98×4= is 392

40,392