Calculating the probability of 31 of 628 items are sampled (no replacement) more than 10x amongst 150 participants drawing 50 items each

Question

I am sorry for having to ask a simple probability question, but I have been thinking about it for weeks and an extensive google search has given no answers.

I have a group of 628 questions. 31 Questions belong to category A. 150 participants will each be given 50 randomly sampled questions from the total 628 questions (no replacement). What is the probability that questions belonging to category A will be answered more than 10 times by the group of participants?

I looked at a binomial distribution using R's binomial density function, but the best I could come up with was the probability of 1 item from category A being administered to the 150 participants if only 1 question was sampled from the total of 628 questions.

    x <- seq(1,50,by=1)
    high.biDen <-dbinom(x,size=150,prob=((31/628)))
    round(high.biDen,2)

    round(sum(high.biDen[10:50]),2)

The answer obtained is .208 or a 20.8% probability that a question from category A will be administered to the participants more than 10 times if 1 question is sampled from the pool of 628 questions. I would like to know the probability if 50 questions are sampled instead of 1.

Is the answer obtained by some sort of bayesian technique? It seems to rely heavily on conditional probability seeing how if one item is sampled only 30 remain to be randomly selected.

Thanks so much for sharing your knowledge and expertise!

-Xander

Answer 1

(1) What is the probability $p_m$ that one participant receives exactly $m$ A-questions? For $m>31$ it is zero, otherwise

$$ p_m = { {31 \choose m} {597 \choose 50-m} \over {628 \choose 50} } $$

(Explanation: we should select $m$ A-questions than $(50-m)$ non-A-questions, and there are ${31 \choose m} {597 \choose 50-m}$ ways to do it.)

(2) What is the probability $q_{k,m}$ that $k$ participants together receive exactly $m$ A-questions? For k>0:

$$ q_{k,m} = \cases {p_m & if $k=1$ \\ \sum_{n=0}^{\min(m,31)} p_n q_{k-1,m-n} & otherwise} $$

(Explanation: let $n$ be the number of A-questions received by the last participant. The probability that he/she received $n$ A-questions, and previous participants received $m-n$ questions, is $p_n q_{k-1,m-n}$. Since we don't know how A-questions he receive, we should sum it by $n$.)