Probability of selecting $n$ orange balls after $k$ steps. This was a problem in my textbook.
Suppose we had a bag with $2$ balls, an orange and a blue ball. If we pick a blue ball, we simply put it back. If we select an orange ball, we put it back but add another orange ball. Suppose we do this $k$ times, what is the probability of all $k$ balls picked are orange?
My work:
We initially have a $0.5$ selecting an orange ball. This is the first step, so the chance of picking an orange ball is $0.5$. If we pick the orange ball, we add another, so the probability of picking another orange ball is $\dfrac{2}{3}$.
I believe we have to take special note of conditional probability, given that you must select an orange ball first before adding a ball.
I'm confused on using conditional probability on which event would be $A$ and which would be $B$. My original intuition was the probability was $\dfrac{k}{k+1}$, for every step $k$ we have $k$ orange balls out of $k+1$ total balls, but I don't think that is right.
 A: Your approach is correct.  Final answer $\prod\limits_{n=1}^k\frac{n}{n+1}=\frac{1}{k+1}$.
A: You're almost right.
Probability of orange first time is $1/2$.
Probability of orange the second time, given that the first was orange, is $2/3$.
And so on.
So the probability of all $n$ balls being orange is
$$ \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1} = \frac{1}{n+1}$$
A: To make your approach rigorous use the Law of Total probability:
$$\mathbb{P}(k\,O)=\mathbb{P}(k\,O:1^{st}O)\mathbb{P}(1^{st}O)+\mathbb{P}(k\,O:1^{st}B)\mathbb{P}(1^{st}B)$$
where ':' denotes 'given', but $\mathbb{P}(k\,O:1^{st}B)$ is obviously zero, and will be each time we apply the Law. At first the probability of choosing orange as you say is $\frac{1}{2}$ so
$$\mathbb{P}(k\,O)=\frac{1}{2}\mathbb{P}(k\,O:1^{st}O)=\frac{1}{2}\mathbb{P}(last(k-1)\,O).$$
The next iteration is
$$\mathbb{P}(last(k-1)\,O)=\mathbb{P}((last(k-1)\,O:2^{nd}O)\mathbb{P}(2^{nd}O)+\mathbb{P}((last(k-1)\,O:2^{nd}B)\mathbb{P}(2^{nd}B)$$
Noting $\mathbb{P}(n^{th}O)=\frac{n}{n+1}$ for each $n$ gives $$ \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{k}{k+1} = \frac{1}{k+1}$$ by telescoping.
A: Probability of first draw being orange ball is $ \ \displaystyle \small \frac{1}{2}$.
Probability of both first and second draws being orange balls is $ \ \displaystyle \small \frac{1}{2} \cdot \frac{2}{3}$
Similarly all $k$ draws being orange balls is $ \ \displaystyle \small \frac{1\cdot 2 \cdot ...k}{2 \cdot 3 \cdot ..(k+1)} = \frac{1}{k+1}$
On your confusion about conditional probability argument, you can look at it this way -
Say $A$ is event of second draw being orange ball and $B$ is event of first draw being orange ball,
Then $\displaystyle \small P(A|B) = \frac{P (A \cap B)}{P(B)} = \frac{(1/2) \cdot (2/3)}{1/2} = \frac{2}{3}$
A: The probability of $k$ orange balls is the probability that the $k+1$ ball is blue, which is $\frac{1}{k+1}$.
