Assigning 40 students to 10 groups of 4 each The scenario given by the problem is as follows:
Suppose a class has forty students, with twenty men and twenty women, including two women,
Alice and Beth, and two men, Chuck and Dan.  All forty students have been randomly assigned
into ten study groups of four students each.
Q1: What is the probability that all women are assigned to groups that include only women?
My ans:
Since there are an equal amount of men and women (20 each), all the women must be assigned to 5 groups of 4, while the men must be assigned to the remaining 5 groups of 4.
For the first group, the probability that we select 4 women is 20C4/40C4. For the second group, the probability that we select 4 women is 16C4/36C4. Repeat this until the fifth group. For the sixth group, the probability that we select 4 men is 20C4/20C4. For the seventh group, the probability that we select 4 men is 16C4/16C4. Repeat this until the tenth group. Multiply all the probabilities together.
There are 10! ways to arrange the 10 groups
10!(((20C4)/(40C4))((16C4)/(36C4))((12C4)/(32C4))((8C4)/(28C4))((4C4)/(24C4))((20C4)/(20C4))((16C4)/(16C4))((12C4)/(12C4))((8C4)/(8C4))*((4C4)/(4C4)))
= 2.63 x 10^-5
Q2: What is the probability that at least one group includes three or more women?
My ans:
The only situation where there are no groups with at least 3 women is when all ten groups have 2 women each.
For the first group, we select 2 women from 20 women and 2 men from 20 men. In the second group, we select 2 women from 18 women and 2 men from 18 men. Repeat this until the tenth group and multiply this together.
The total number of possibilities is choose 4 from 40 for the first group, then choose 4 from 36 for the second group, etc.
Subtract the probability of this unique situation from 1
1 - ((20C2)(18C2)(16C2)(14C2)(12C2)(10C2)(8C2)(6C2)(4C2)(2C2))^2/((40C4)(36C4)(32C4)(28C4)(24C4)(20C4)(16C4)(12C4)(8C4)(4C4))
=0.99956
Would really appreciate it if someone can verify whether my reasoning and answers are correct. Thank you for your help in advance.
 A: Think of $10\times4=40$ numbered open spots subdivided in $10$ groups of $4$ spots  where persons will be placed.
There are $\binom{40}{20}$ ways to select $20$ of these spots, so if this is done then the probability that the women end up on these selected spots is $\binom{40}{20}^{-1}$.

Q1
The answer should be: $$\binom{10}5\binom{40}{20}^{-1}$$
This because there are $\binom{10}5$ ways to select $20$ spots in such a way that they are the spots in exactly $5$ groups. Actually this boils down to selecting $5$ of $10$ groups.
Q2
The answer should be:$$1-\binom42^{10}\binom{40}{20}^{-1}$$
This because there are $\binom42^{10}$ ways to select $20$ spots in such a way that from every group $2$ spots are selected.

Simplify your answers (just write them out and defactorize) and compare.
A: Imagine the groups having existing positions waiting to be filled with either male/female people. The total number of assignments of sex to positions is then $40\choose 20$ while the number of assignments of sex to groups is $10\choose 5$. The ratio $1.83\cdot 10^{-9}$ is the probability that the groups are single sex, i.e. the answer to Q1.
The negation of the probability in Q2 we have that every group contains at most 2, and therefore exactly 2, women. The number of assignments here is ${4 \choose 2}^{10}$ while the total is $40\choose 20$ again, giving $4.39\cdot 10^{-4}$. So, the answer here is $0.99956$.
