A student has a 30% chance to get a specific question correct. In a class of 30, given that the probability that at least 6 students in the class can do the question is 0.923.
What is the probability that only 2 students among the first 8 selected students in that class could do the question given that at least 6 students in the class can do the question?
I'm having really big problems because of how the 2 scenarios are not independent of each other, so I'm getting stuck at that part. I am unable to make a binomial model considering that at least 6 students did the question as well. The answer key states that the answer is 0.299.
 A: Given the first sentence (after your title) it seems reasonable to assume that the problem intends for the number of students to get the problem right to be binomially distributed.
Indeed, when checking, we have $\sum\limits_{n=6}^{30}\binom{30}{n}0.3^n\times 0.7^{30-n}\approx 0.923405248$ which matches with the problem statement.
Under this assumption then, we ask the question, "given that at least six students know the answer, what is the probability that among the first eight students exactly two know the answer."
For this, remember that $Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$.  We already know the probability that at least six students know the answer as it was given in the problem statement (and we verified it above).  What remains is to calculate the probability that simultaneously we have exactly two of the first eight students knowing the answer while at the same time having at least six total students of the 30 knowing the answer.
To calculate this, we know that among the first $8$ students, exactly two know the answer.  This occurs with probability $\binom{8}{2}0.3^2\times 0.7^6$.  Then, among the remaining $22$ students, at least four know the answer.  This occurs with probability $\sum\limits_{n=4}^{22}\binom{22}{n}0.3^n\times 0.7^{22-n}$.
Completing the arithmetic and multiplying the results will then tell you the probability that both occur simultaneously.  Dividing this result by the earlier result will give you the conditional probability.
wolfram calculations
A: Let´s define some events:
$A$: 2 students among the first 8 selected students in that class could do the question
$B$: At least 6 students in the class can do the question
Applying Bayes´ Theorem
$$P(A|B)=\frac{P(B|A)\cdot P(B)}{P(A)}$$
$P(B|A)$ is the probability that at least $4 (=6-2)$ students of $22$ $(=30-8)$ can do the question: 
$$\sum_{k=4}^{22}\binom{22}{k}\cdot 0.3^k\cdot 0.7^{22-k}$$
Is this comprehensible? If yes, then I think you can calculate $P(A)$ and $P(B)$
