5
$\begingroup$

Before I proceed with my queries I think it's best to present the question at hand.

A class consisting of 4 males and 12 females in randomly divided into 4 groups of 4. What is the probability that each group consists of a male and three females?


Now the given answer for this was $\frac{64}{455}$.

The closest I could possibly get to that was $455$.

Whenever I come across these types of questions I generally categorise it like so:

  • Does order matter?
    • Yes: then it is a permutation (% increases)
    • No: then it is a combination (% decreases)
  • Does it allow repetitions? - Make use of respective formulae

So reading the question I will immediately think, "No, order does not matter" which I find a bit hard to explain but Male Student A may be in the first group and Male Student B in the second group OR Vice-Versa (hence why I would consider it a combination..).

So once that is out of the way, I proceed and think, "Yes, there may be repetitions". For consistencys sake, Male Student A may be in the first group twice.

Now that I got that chunk out of the way I pray to the seven Gods that I got the previous two right (even though it's purely logic, certain question try to trick you), and hence begin working.


Pardon if the following notation is incorrect: $$C\binom{16}{4}$$

I tried the following: $$\therefore \frac{16!}{12!\times 4!}\div 4 = 455$$

And then: $$\therefore \frac{16!}{12!\times 4!}\div 4! = \frac{6}{455}$$

However the answer presented in my answer sheet is: $$\frac{64}{455}$$

$\endgroup$
5
$\begingroup$

Let's look at the combinations, and we'll do it one group at a time. We are looking for the probability that each of the four groups has exactly one boy and three girls.

For the first group, we are choosing $4$ people out of $16$. One is a boy, out of $4$ possibilities. Three are girls, out of $12$ possibilities. The choices of boy and girls are independent, so we can multiply their counts. Therefore, the probability that the first group has one boy and three girls is

$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}}=\frac{44}{91}$$

Now for the second group. After the first group is chosen suitably, we have to choose $1$ boy out of $3$ remaining and $3$ girls out of $9$ remaining. But the total number of possible groups is $4$ people out of $12$ remaining. Therefore, given that the first group was successful, the probability that the second group is successfully chosen is

$$\frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}}=\frac{28}{55}$$

Now for the third group. After the first two groups are chosen suitably, we have to choose $1$ boy out of $2$ remaining and $3$ girls out of $6$ remaining. But the total number of possible groups is $4$ people out of $8$ remaining. Therefore, given that the first and second groups were successful, the probability that the third group is successfully chosen is

$$\frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}}=\frac{4}{7}$$

If the first three groups are correct, so is the fourth. Therefore, the total probability of success with all four groups is

$$\frac{44}{91}\cdot\frac{28}{55}\cdot\frac{4}{7}=\frac{64}{455}$$

Or, if you want the final probability as one large calculation,

$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}} \cdot \frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}} \cdot \frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}} =\frac{64}{455}$$

$\endgroup$
1
  • $\begingroup$ Such a great answer. Thanks for taking the time to explain all of this. Makes perfect sense. $\endgroup$
    – Juxhin
    Apr 26 '15 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.