Balls are randomly taken out of a box of 4 red and 6 green balls, without replacement, until we have a ball of each colour. What is the probability of

*

*Needing exactly two draws

*Needing two or three draws

*Needing to make exactly three draws if the first two balls drawn are of the same colour

For part 1. I computed that the number of ways we can choose a red followed by a green or vice versa is $|A|=6\cdot4$. The sample space itself has $10\choose2$ items so $$P(A)=\frac{6\cdot4}{10\choose2}=\frac{8}{15}$$ Part 2. and 3. posed the same issue for me. I approached part 2. with two events $A$: It takes two throws, $B$: It takes three throws. Then those two events are mutually exclusive so $P(A\cup B)=P(A)+P(B)$ which we are looking for. Here is the issue for 2. and 3.. I don't know how to calculate the probability that the first two are of the same colour. I thought that it may be $$\frac{{6\choose4}+{4\choose2}}{10\choose2}$$ But even if this were right I don't know how to mathematically express the fact that the 3rd one must be different. My theory is that (assuming the above is correct) it may be $$\frac{{6\choose4}\cdot{4\choose1}+{4\choose2}\cdot{6\choose 1}}{10\choose3}$$
EDIT 1
I got solutions for part 1. and 3. but am still wrapping my head around 2. For 1. you have the event $A$: The balls are of different colours. For $A$ we have $|A|=6\cdot4+4\cdot6$ and a sample space $|\Omega|=10\cdot9$ hence $P(A)=\frac{8}{15}$ similar to the above. For 3. we need to find $P(B|C)$. We know that $$P(B|C)=\frac{P(B\cap C)}{P(C)}$$. Furthermore $P(B\cap C)=P(B)$ because $B\cap C$ is the event that three balls are chosen. If three balls are chosen then the first two must have been the same colour. Hence $$P(B|C)=\frac{P(B)}{P(C)}=\frac47$$
EDIT 2
Question 2. has been resolved as well. Given that all of the events involved are pairwise disjoint the answer is just the sum of the prob that two are drawn and three are drawn. $$P(A)+P(B)=\frac{12}{15}$$
 A: To need exactly $n$ draws is the event for obtaining $n-1$ consecutive draws of one colour followed by one ball of the other.
$$\dfrac{\dbinom 4{n-1}\dbinom 61+\dbinom 6{n-1}\dbinom 41}{\dbinom{10}{n-1}\dbinom {9-n}1}\tag{$n\in\{2,3,4,5,6,7\}$}$$
Note: There is importance in distinguishing the earlier draws from the possible terminal draw.

The probability for needing exactly two draws to obtain different colored balls, when drawing without replacement from four red and six green balls is: the probability for obtaining one balls of one colour then one ball of the other, when drawing a ball then another.
$$\dfrac{\dbinom 41\dbinom 61+\dbinom 61\dbinom 41}{\dbinom{10}1\dbinom 91}=\dfrac{4\cdot 6+6\cdot 4}{10\cdot 9}$$
Which is coincidentally equal to $\binom 41\binom 61/\binom{10}2$.
Likewise:
The probability for needing exactly three draws to obtain different colored balls, when drawing without replacement from four red and six green balls is: the probability for obtaining two balls of one colour then one ball of the other, when drawing two balls then a third.
$$\dfrac{\dbinom 42\dbinom 61+\dbinom 62\dbinom 41}{\dbinom{10}2\dbinom{8}1}=\dfrac{4\cdot 3\cdot 6+6\cdot 5\cdot 4}{10\cdot 9\cdot 8}$$

But even if this were right I don't know how to mathematically express the fact that the 3rd one must be different.

That was how.  The denominator may be any of the following:
$$\binom{10}3\binom 31=\binom{10}2\binom 81=\binom {10}1\binom 92$$

*

*Select 3 from 10 balls, and 1 of those 3 (to be the last).

*Select 2 from 10 balls, then 1 from the remaining 8.

*Select 1 from 10 balls (to be the third), then 2 from the remaining 9 (to be before it).


So the probability for needing two or three draws is: ...


The probability that exactly three draws are needed when given that the first two draws are of the same colour has the form:
$$\dfrac{\mathsf P(R_1,R_2,G_3)+\mathsf P(G_1,G_2,R_3)}{\mathsf P(R_1,R_2)+\mathsf P(G_1,G_2)}=\dfrac{\lvert R_1,R_2,G_3\rvert+\lvert G_1,G_2,R_3\rvert}{\lvert R_1,R_2,(R_3\cup G_3)\rvert+\lvert G_1,G_2,(R_3\cup G_3)\rvert}$$
