Probability of future success when trials are dependent: In 4 draws w/o replacement, what is the probability that draws 3 & 4 will be successful?

Question

Suppose that I have a bag with 14 marbles, 9 green and 5 red.
Person A will first draw 2 marbles and then Person B will draw 2 marbles.
Can I determine the probability that person B will draw 2 green marbles before I let person A draw?

Edit: The order of events are:

1. Determine the probability that Person B will draw 2 green marbles.
2. Person A draws 2 marbles (unknown color)
3. Person B draws 2 marbles

I understand that using hypergeometric probability I can easily find the probability that person A will draw 2 green marbles

$$\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}.$$

K = 9 ; k = 2 ; N = 14 ; n = 2

however if I only care about person B drawing 2 green marbles, it seems to me that now n = 4 but k could be k = 2, k = 3, or k = 4 but it doesn't matter so long as the last 2 instances of n yield k=2.

Am I thinking about this in the right way? Any help is appreciated

Answer 1

Consider that there are $9$ green marbles and $5$ red marbles in the beginning. We split cases:

Case 1: $A$ draws $2$ green marbles.

The probability of this case is $\binom92/\binom{14}2=\frac{36}{91}$.

After $A$ draws, there would remain $7$ green marbles, $12$ marbles in total.

The probability of $B$ drawing $2$ green marbles after that is $\binom72/\binom{12}2=\frac{21}{66}$.

So, the probability of $B$ drawing $2$ green marbles through this case is $$\frac{36}{91}\cdot\frac{21}{66}\text.$$

Case 2: $A$ draws $1$ green marble and $1$ red marble.

The probability of this case is $\binom91\binom51/\binom{14}2=\frac{45}{91}$.

After $A$ draws, there would remain $8$ green marbles, $12$ marbles in total.

The probability of $B$ drawing $2$ green marbles after that is $\binom82/\binom{12}2=\frac{28}{66}$.

So, the probability of $B$ drawing $2$ green marbles through this case is $$\frac{45}{91}\cdot\frac{28}{66}\text.$$

Case 3: $A$ draws $2$ red marbles.

The probability of this case is $\binom52/\binom{14}2=\frac{10}{91}$.

After $A$ draws, there would remain $9$ green marbles, $12$ marbles in total.

The probability of $B$ drawing $2$ green marbles after that is $\binom92/\binom{12}2=\frac{36}{66}$.

So, the probability of $B$ drawing $2$ green marbles through this case is $$\frac{10}{91}\cdot\frac{36}{66}\text.$$

Total probability

So, across all cases, the probability is $$\begin{align} &\left(\frac{36}{91}\cdot\frac{21}{66} +\frac{45}{91}\cdot\frac{28}{66} +\frac{10}{91}\cdot\frac{36}{66}\right)\\ =&\frac{36\cdot21+45\cdot28+10\cdot36}{91\cdot66}\\ =&\frac{756+1260+360}{6006}\\ =&\frac{2376}{6006}\\ =&\frac{36}{91}\text. \end{align}$$