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
Answer:
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.
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.