Two parents are carriers for Galactosemia, if they have 4 kids, what is the probability that AT LEAST one child will have the disease? I used the indirect method 1-(3/4)^4 and got 175/256, but how can I go about using the direct method?
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Your answer is correct. Using an indirect method, which I find to be the easiest, you can determine the probability that none of the children have Galactosemia and then the probability that at least one of the children has Galactosemia is 1 - the probability that none of the children have Galactosemia. Just to fill out the explanation in case someone else reads this, we know that the probability of any one child having Galactosemia is 1/4, and we know that the probability that any one child can be a carrier, but not have Galactosemia is 1/2, and we know that the probability that any one child is not a carrier and thus won't have Galactosemia is 1/4. So, the probability that a child will have Galactosemia is 1/4 and the probability that they won't have it is 3/4. So, the probability that none of the children have Galactosemia is 3/4 x 3/4 x 3/4 x 3/4 = 81/256. So, the probability that at least one of the children has Galactosemia is 1 - 81/256 = 256/256 - 81/256 = 175/256.
The direct method would be to look at the probabilities of all of the combinations. The probability that all of the kids have Galactosemia is (1/4)^4 = 1/256. The probability that three of the kids have it and one does not is (1/4 x 1/4 x 1/4 x 3/4) = 3/256, plus the three other combinations (1/4 x 1/4 x 3/4 x 1/4), (1/4 x 3/4 x 1/4 x 1/4), (3/4 x 1/4 x 1/4 x 1/4). The probability that 2 of the kids have it and the other two do not is (1/4 x 1/4 x 3/4 x 3/4) = 9/256, plus the five other combinations (3/4 x 1/4 x 1/4 x 3/4), (3/4 x 3/4 x 1/4 x 1/4), (3/4 x 1/4 x 3/4 x 1/4), (1/4 x 3/4 x 1/4 x 3/4), (1/4 x 3/4 x 3/4 x 1/4). The probability that one of the kids has it and the other three do not is (1/4 x 3/4 x 3/4 x 3/4) = 27/256, plus the three other combinations (3/4 x 3/4 x 3/4 x 1/4), (3/4 x 3/4 x 1/4 x 3/4), (3/4 x 1/4 x 3/4 x 3/4). So, the total probability that at least one of the children has Galactosemia would be: 1/256 + (4 x 3/256) + (6 x 9/256) + (4 x 27/256) = 1/256 + 12/256 + 54/256 + 108/256 = 175/256
Using the combinations formula n!/(n-r)!r! you can find the multiplier for each of these: 4 things taken 1 at a time = 4 combinations, 4 things taken 2 at a time = 6 combinations and 4 things taken 3 at a time = 4 combinations.
The direct method would be to look at the probabilities of all of the combinations. The probability that all of the kids have Galactosemia is (1/4)^4 = 1/256. The probability that three of the kids have it and one does not is (1/4 x 1/4 x 1/4 x 3/4) = 3/256, plus the three other combinations (1/4 x 1/4 x 3/4 x 1/4), (1/4 x 3/4 x 1/4 x 1/4), (3/4 x 1/4 x 1/4 x 1/4). The probability that 2 of the kids have it and the other two do not is (1/4 x 1/4 x 3/4 x 3/4) = 9/256, plus the five other combinations (3/4 x 1/4 x 1/4 x 3/4), (3/4 x 3/4 x 1/4 x 1/4), (3/4 x 1/4 x 3/4 x 1/4), (1/4 x 3/4 x 1/4 x 3/4), (1/4 x 3/4 x 3/4 x 1/4). The probability that one of the kids has it and the other three do not is (1/4 x 3/4 x 3/4 x 3/4) = 27/256, plus the three other combinations (3/4 x 3/4 x 3/4 x 1/4), (3/4 x 3/4 x 1/4 x 3/4), (3/4 x 1/4 x 3/4 x 3/4). So, the total probability that at least one of the children has Galactosemia would be: 1/256 + (4 x 3/256) + (6 x 9/256) + (4 x 27/256) = 1/256 + 12/256 + 54/256 + 108/256 = 175/256
Using the combinations formula n!/(n-r)!r! you can find the multiplier for each of these: 4 things taken 1 at a time = 4 combinations, 4 things taken 2 at a time = 6 combinations and 4 things taken 3 at a time = 4 combinations.