An urn has $4$ black, $5$ white balls. A ball is drawn, the color is noted, and then the ball is returned back to the urn.
I would like to verify my answers:
a) What is the probability that 5 black balls are observed in $15$ draws?
Since it is replaced, we can think of it as: I want $5$ black balls, $10$ white balls, in any order:
$$\binom{15}{5}\left(\frac{4}{9}\right)^5\left(\frac{5}{9}\right)^{10}$$
b) What is the probability that $15$ draws are required until the first black ball comes.
This wording confuses me. Am I to assume there are a total of $16$ draws then, which the $16$th one being the first black?
$$\left(\dfrac{5}{9}\right)^{15}\left(\dfrac{4}{9}\right)$
In this, there is only one way this can happen (first $15$ are white, last is black)
c) What is the probability that $15$ draws are required until the fifth black ball comes?
Once again, am I to assume that on the $16$th draw, the fifth black ball will come?
In that sense I can have $15$ spots to store $4$ blacks, and $11$ whites.
Thus, a possible sequence is $BBBBWWWWWWWWWWWB$
$$\binom{15}{4}\left(\frac{4}{9}\right)^4\left(\frac{5}{9}\right)^{11}\left(\frac{4}{9}\right)$$
Is this correct? Also, I am learning about distributions right now, and was wondering if this problem can be done using a negative binomial distribution because it follows similar logic about $x$ amount of failures before the $r^{th}$ success?