There are 25 marbles in an opaque jar: 7 red, 8 blue, 6 yellow and 4 green. Four marbles are chosen at random without replacement. What is the probability of choosing green, then red, then a blue, and then blue, in that order. Answer in reduced fraction form. Do not write a decimal answer.

Step 1 ：There are 25 marbles in total: 7 red, 8 blue, 6 yellow and 4 green. We are to find the probability of choosing a green marble, then a red marble, then a blue marble, and then another blue marble, in that order.

Step 2 ：The total number of marbles is 25. The number of green marbles is 4, so the probability of choosing a green marble first is \(\frac{4}{25}\).

Step 3 ：After choosing a green marble, there are 24 marbles left. The number of red marbles is 7, so the probability of choosing a red marble next is \(\frac{7}{24}\).

Step 4 ：After choosing a green and a red marble, there are 23 marbles left. The number of blue marbles is 8, so the probability of choosing a blue marble next is \(\frac{8}{23}\).

Step 5 ：After choosing a green, a red, and a blue marble, there are 22 marbles left. The number of blue marbles is now 7 (since we already chose one), so the probability of choosing another blue marble is \(\frac{7}{22}\).

Step 6 ：The probability of all these events happening in order is the product of their individual probabilities. Therefore, the probability is \(\frac{4}{25} \times \frac{7}{24} \times \frac{8}{23} \times \frac{7}{22} = \frac{98}{18975}\).

Step 7 ：Final Answer: The probability of choosing a green marble, then a red marble, then a blue marble, and then another blue marble, in that order, is \(\boxed{\frac{98}{18975}}\).