A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag   contains    7   blue    balls   and 3   green   balls.
I am not sure about my answer but please allow me to share.

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*If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?

There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.
Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.
Hence,
$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$
The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.


*If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?

$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$


*If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?

Work it out first where the order they are drawn is important.
$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$
Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is
$20 × (\frac{1}{24}) = (\frac{5}{6})$
Any comments or suggestions will be much appreciated. Thank you in advance.
 A: 
If five balls are drawn in succession at random without replacement, what is the probability of drawing one green, one blue, one green, one blue in order?

You have correctly calculated that the probability that the sequence GBGB occurs in the first four draws is
$$\Pr(GBGB) = \frac{3}{10} \cdot \frac{7}{9} \cdot \frac{2}{8} \cdot \frac{6}{7}$$
However, it could also happen in the second four draws after being preceded by either a green or a blue ball.
$$\Pr(GGBGB) = \frac{3}{10} \cdot \frac{2}{9} \cdot \frac{7}{8} \cdot \frac{1}{7} \cdot \frac{6}{6}$$
and
$$\Pr(BGBGB) = \frac{7}{10} \cdot \frac{3}{9} \cdot \frac{6}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}$$
Since these three cases are mutually exclusive and exhaustive, the probability that the sequence GBGB appears in the first five draws is found by adding the above probabilities.

If three balls are drawn at the same time, what is the probability that all three balls drawn are blue?

Your answer is correct.
An alternate method is to observe that we can select three of the ten balls in
$$\binom{10}{3}$$
ways.  Of these, we can select three blue balls in
$$\binom{7}{3}$$
ways.  Hence, the probability that a random selection of three balls from a bag containing seven blue and three green balls results in the selection of three blue balls is
$$\Pr(\text{three blue}) = \frac{\dbinom{7}{3}}{\dbinom{10}{3}}$$

If five balls are drawn at the same time, what is the probability that three balls are blue and two balls are green.

There are
$$\binom{10}{5}$$
ways to select five of the ten balls.  Of these, we can select three of the seven blue balls and two of the three green balls in
$$\binom{7}{3}\binom{3}{2}$$
ways.  Hence, the probability that a random selection of five balls from a bag with seven blue and three green balls results in the selection of three blue and two green balls is
$$\Pr(\text{three blue and two green balls}) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$
Your answer resulted in too large a probability since the number of ordered sequences of three blue and two green balls is
$$\binom{5}{2}$$
since we must choose two of the five positions in the sequence for the green balls.  Notice that
$$\binom{5}{2}\left(\frac{7}{10}\right)\left(\frac{6}{9}\right)\left(\frac{5}{8}\right)\left(\frac{3}{7}\right)\left(\frac{2}{6}\right) = \frac{\dbinom{7}{3}\dbinom{3}{2}}{\dbinom{10}{5}}$$
A: Problem 1.
Your answer to problem 1. is wrong, because you have computed the probability
of getting G-B-G-B in the first four drawings.  The problem specifically
asked for the probability of the satisfying sequence occuring anywhere
within the first five drawings.
The actual problem is much more complicated, and is best attacked by the
following approach:  the answer will be
$$\frac{N\text{(umerator)}}{D\text{(enoninator)}}$$
where
$$D = 10 \times 9 \times 8 \times 7 \times 6.$$
Because of the constraints of the problem, the easiest approach is to identify
each of the 4 possible satisfying sequences, and enumerate each possibility.
$N_1~: G-B-G-B-G ~:~N_1 = 3 \times 7 \times 2 \times 6 \times 1.$
$N_2~: G-B-G-B-B ~:~N_2 = 3 \times 7 \times 2 \times 6 \times 5.$
$N_3~: G-G-B-G-B ~:~N_3 = 3 \times 2 \times 7 \times 1 \times 6.$
$N_4~: B-G-B-G-B ~:~N_4 = 7 \times 3 \times 6 \times 2 \times 5.$
Final answer:
$$ \frac{N_1 + N_2 + N_3 + N_4}{D}.$$

Problem 2.
Your answer is correct.

Problem 3.
Your answer is almost right.  You overlooked overcounting.
There are only $\binom{5}{2} = 10$ ways of selecting which slots the 2 blue
balls will go in.  So you need to apply a scaling factor of (1/2) to your answer.
