I'm struggling to understand the procedure to solve the following problem:
The group consists of 9 people, 3 men and 6 women. They are divided randomly into 3 groups (3 persons each). What's the probability that..
a) all men are in the same group? b) all men are in different groups?
I somehow managed to calculate a) as follows:
all combinations: 9!/(3!^4) = 280
combinations of all men in one group: 6!/(3!*3!*2!) = 10
probability: 10/280 = 0,03571..
Please, do let me know if I am doing this wrong. Also, I do not know how to figure out b) all men in different groups. Asking for your advice, thank you!
$\endgroup$ 22 Answers
$\begingroup$I think this can be done in a straight-forward way. The order in which groups are formed does not matter, so we can assume that the process begins by randomly drawing people from the pool to add to the first group. This makes it easier because now you simply have to compute {M is drawn} * {M is drawn again} * {M is drawn a third time}. This is
$$\frac39 \cdot \frac28 \cdot \frac17 = \frac1{84}$$
Then the probability that any group consists of only men is 3 times that, so $\frac{1}{28}$. Alternatively, begin after the first man has been assigned, and just multiply $\frac28 \cdot \frac17 = \frac{1}{28}$.
For the second, again we can just fix one order. Assume we first draw people for just group one, then the probability to have exactly one man in there (which is necessary to have one men in each group) is
$$3 \cdot \frac39 \cdot \frac68 \cdot \frac57$$
(Compute the probability for MWW, then multiply it by 3 to account for the permutations WMW and WWM.) Similarly for group two:
$$3 \cdot \frac26 \cdot \frac45 \cdot \frac34$$
and group three:
$$3 \cdot \frac13 \cdot \frac22 \cdot \frac11$$
This one just yields 1, which makes sense because after the first two groups have one man in each, the last one will inevitably also have one.
In total, we get
$$3 \cdot \frac39 \cdot \frac68 \cdot \frac57 \cdot 3 \cdot \frac26 \cdot \frac45 \cdot \frac34 = \frac9{28}$$
$\endgroup$ 1 $\begingroup$First of all one should determine the meaning of "they are divided randomly into 3 groups (3 persons each)."
I would assume that the procedure is following: there are 9 positions separated in the groups of the size 3: $(123)(456)(789)$ and the people are randomly distributed between the positions.
Then
there are altogether $9!$ ways to distribute the people.
there are $3\times 3!\times 6!$ ways to distribute the people in such a way that all men are in the same group.
Explanation: $3$ ways to choose the group, $3!$ ways to permute the men inside the group, $6!$ ways to place women on remaining positions.
- there are $3!\times 3^3\times 6!$ ways to distribute the people in such a way that all men are in different groups.
$\endgroup$ 2Explanation: $3!$ ways to permute the men between the groups, $3$ ways to choose the position of each men inside his group, $6!$ ways to place women on remaining positions.
