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"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"

$left(dfrac{6}{10}right)left(dfrac{5}{9}right)left(dfrac{4}{8}right)$

So why can't we use that logic to answer this question?

"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"

My answer: $left(dfrac{4}{11}right)left(dfrac{3}{10}right)left(dfrac{5}{9}right)left(dfrac{4}{8}right)left(dfrac{2}{7}right)$

But the correct answer is $dfrac{(_4C_2) cdot (_5C_2) cdot (_2C_1)}{_{11}C_5}$ (where $C$ is a combination).

Why doesn't the logic from the first problem work here?

The draws are without replacement in all cases.

What you calculated is the probability of selecting two red marbles, two blue marbles, and one green marble in that order. However, if we select blue, green, red, red, blue, we still get two red marbles, two blue marbles, and one green marble.

To correct your attempt, we must multiply by the number of orders in which we could get two red marbles, two blue marbles, and one green marble. Choose two of the five positions for the blue marbles and two of the remaining three positions for the green marbles. The only green marble must go in the remaining position. Then we obtain
$$binom{5}{2}binom{3}{2}binom{1}{1}left(frac{4}{11}right)left(frac{3}{10}right)left(frac{5}{9}right)left(frac{4}{8}right)left(frac{2}{7}right) = frac{20}{77}$$

Since we do not care about the order in which the marbles are selected, it is simpler to calculate the probability of selecting two of the four red marbles, two of the five blue marbles, and one of the two green marbles when we select five of the eleven marbles, which yields
$$frac{dbinom{4}{2}dbinom{5}{2}dbinom{2}{1}}{dbinom{11}{5}} = frac{20}{77}$$

"So why can't we use that logic to answer this question?"

Using the logic in the first question you actually calculate the event that "the first chosen student is a girl, the second is a girl and the third is a girl". This is exactly the same event as "all $3$ students chosen will be a girl".

Using the logic in the second situation you similarly calculate the event that "the first chosen marble is red, the second is red, the third is blue, the fourth is blue and the fifth is green". This is definitely not the same event as "$2$ chosen marbles are red, $2$ are blue and $1$ is green".

If that event occurs then it is not excluded that e.g. the first chosen marble is green (hence not red).

You can make use of the method, but then must not forget that the answer found at first hand must be multiplied by the number of orders that are possible.

The first situation with the students is simpler than the second situation with the marbles because we are interested in only one kind of studens (namely girls). So we don't have to deal with different kinds of students.

This gives the simpler answer for the students-question, which doesn' work for the marbles-question.

We can apply the more general logic from the marbles-question to the students question and get as answer:

(6C3) / (10C3) = (6 * 5 * 4) / (10 * 9 * 8)

which is equal to the simpler formula you gave.

The first question solved in the second method looks as follows:
$$frac{{6choose 3}}{{10choose 3}}=frac{6cdot 5cdot 4}{10cdot 9cdot 8}$$
Interpretation: There are ${6choose 3}$ ways to choose $3$ girls out of $6$ and there are ${10choose 3}$ ways to choose $3$ students out of $10$, hence the probability is the ratio of number of favorable outcomes to the total number of possible outcomes.

Now compare it with the answer of the second problem and try to interpret the selections.