Final answer:
The probability of selecting 4 genes without replacement from the 12 genes, making sure none come from the same pair, is approximately 0.03. This is calculated by the ratio of the combination of choosing 4 pairs out of 6 to the combination of choosing any 4 genes out of 12, which does not match any of the provided multiple-choice options.
Step-by-step explanation:
To calculate the probability that we select four genes, none of which are from the same pair, when sampling without replacement from the 12 genes, we can use the concept of permutations without replacement. Since each pair of genes cannot be selected more than once, we have a limitation on our selection.
Firstly, there are 6 pairs of alleles, giving us 6 choices for the first gene. Having chosen one, we cannot choose the same pair again, so we have 5 choices left for the second gene, 4 for the third, and 3 for the fourth. Calculating the total number of ways we can pick 4 different pairs out of 6 without regard to order gives us a calculation using combinations (denoted as C(n, k) where n is the total number of items to choose from, and k is the number of items to be chosen):
C(6, 4) = 6! / (4!(6 - 4)!) = 15
Next, we know that there are a total of 12 genes to choose from, and we are choosing 4. So, the total number of ways to choose 4 genes out of 12 is:
C(12, 4) = 12! / (4!(12 - 4)!) = 495
Now the probability is the number of favorable outcomes over the total number of possible outcomes:
P = favorable outcomes / total outcomes = C(6, 4) / C(12, 4)
P = 15 / 495 = 0.03030303, or about 0.03 when rounded to two decimal places.
As none of the provided options A) 0.134 B) 0.209 C) 0.288 D) 0.421 E) 0.521 match this calculated value, it seems there may be an error in the question or options provided.