Probability of more than 13 balls remaining in the box after some selections Question
A box contains 20 balls, of which exactly 8 are red and 12 are green. Two balls are randomly selected from the box, without replacement. Then, any red balls selected are put back into the box, while any green balls selected are thrown away. Finally, 5 balls are randomly selected from the box, without replacement. Suppose that, following this procedure, all 5 balls are green. Find the probability that there remains more than 13 balls in the box.
My working
Observe that the only case when there are less than 14 balls in the box is when the two balls that are selected initially are both green, so we have three cases to consider here.
Case 1: Choose 2 red balls, put them back, then choose 5 green balls.
Case 2: Choose 1 red ball, then 1 green ball. Put the red ball back, throw the green one away and then choose 5 green balls.
Case 3: Choose 1 green ball, then 1 red ball. Put the red ball back, throw the green one away and then choose 5 green balls.
$\mathbb{P} (\mathrm {Case\ 1}) = \frac 8 {20} * \frac 7 {19} * \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} = \frac {231} {30685}$
$\mathbb{P} (\mathrm {Case\ 2}) = \frac 8 {20} * \frac {12} {19} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} = \frac {308} {30685}$
$\mathbb{P} (\mathrm {Case\ 3}) = \mathbb{P} (\mathrm {Case\ 2})$
$\therefore \mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {231} {30685} + 2 * \frac {308} {30685} = \frac {847} {30685}$
Answer
$\mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {154} {211}$

My answer is obviously off by a lot ): Am I missing cases, or is my thinking too naive etc.? I am not sure where I have gone wrong, so any intuitive explanations about my mistake(s) and also what the correct solution should be will be greatly appreciated!
P.S. My professor did provide a solution, but I am unable to understand his way of thinking, so I thought posting the problem here and asking for ideas would be better.
 A: I haven't checked your arithmetic, but it appears that you aren't computing the conditional probability.  Let $G$ be the event that all five balls are green, and let $T$ be the event that there are more than $13$ balls in the box.  We are asked for $$\Pr(T|G)=\frac{\Pr(T\cap G)}{\Pr(G)}$$
It looks to me like you've only computed the numerator.  You need to compute the probability that the first two balls are green and the last five are green, and add it to the numerator, in order to get the denominator.
A: So, based on another answer (it was more of a hint), it seems that I have simply forgotten to consider the conditional probability aspect of the problem.
We have indeed 3 cases that fit our criteria, but that is actually out of a total of 4 cases, the fourth being that the first two balls chosen are both green. I conveniently left the fourth case out of my original working and considering it will get us the correct answer, as shown below.
Case 1: Choose 2 red balls, put them back, then choose 5 green balls.
Case 2: Choose 1 red ball, then 1 green ball. Put the red ball back, throw the green one away and then choose 5 green balls.
Case 3: Choose 1 green ball, then 1 red ball. Put the red ball back, throw the green one away and then choose 5 green balls.
Case 4: Choose 2 green balls, throw them away and then choose 5 green balls, which is equivalent to simply choosing 7 green balls.
$\mathbb{P} (\mathrm {Case\ 1}) = \frac 8 {20} * \frac 7 {19} * \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} = \frac {231} {30685}$
$\mathbb{P} (\mathrm {Case\ 2}) = \frac 8 {20} * \frac {12} {19} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} = \frac {308} {30685}$
$\mathbb{P} (\mathrm {Case\ 3}) = \mathbb{P} (\mathrm {Case\ 2})$
$\mathbb{P} (\mathrm {Case\ 4}) = \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} * \frac 6 {14} = \frac {33} {3230}$
$\therefore \mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {\mathbb{P} (\mathrm {Case\ 1}) + \mathbb{P} (\mathrm {Case\ 2}) + \mathbb{P} (\mathrm {Case\ 3})} {\mathbb{P} (\mathrm {Case\ 1}) + \mathbb{P} (\mathrm {Case\ 2}) + \mathbb{P} (\mathrm {Case\ 3}) + \mathbb{P} (\mathrm {Case\ 4})} = \frac {\frac {231} {30685} + 2 * \frac {308} {30685}} {\frac {231} {30685} + 2 * \frac {308} {30685} + \frac {33} {3230}} = \frac {154} {211}$
