Q:

Draw balls in the urn Probabilities

An urn contains $7$ balls, $4$ of which are black and the rest are white.

Starting with all $7$ balls in the urn, balls are drawn randomly with replacement.
Find the probability that among the first $5$ balls which are drawn:
a. Exactly $2$ of them are white.
b. At least $3$ of them are black.

For 1 a) I used Binomial Distribution:
$$\binom52\cdot\left(\frac37\right)^2\cdot\left(\frac47\right)^3$$
For 1 b) I used again Binomial distribution but because it is at least $3$ of them are black I did:
$$P(X=3)+P(X=4)=\binom53\cdot\left(\frac47\right)^3\cdot\left(\frac37\right)^2+\binom54\cdot\left(\frac47\right)^4\cdot\left(\frac37\right)^1=0.5711$$
Am I doing it right?

2) Starting with all $7$ balls in the urn, balls are drawn randomly without replacement until all $3$ white balls are found. What is the probability that the third white ball is found on the 
a. fourth attempt? b. seventh attempt? 
For $2.a$  I did: 
The combinations can be:
$wwbw$
$P =\dfrac37\cdot\dfrac26\cdot\dfrac45\times\dfrac14 =\dfrac1{35}$
$wbww$ 
$P =\dfrac37\cdot\dfrac46\cdot\dfrac25\times\dfrac14 = \dfrac1{35}$
$bwww$ 
$P = \dfrac47\cdot\dfrac36\cdot\dfrac25\times\dfrac14 = \dfrac1{35}$
so : $P = \dfrac3{35}$, right?
for question $2 b)$ is there a way to do this without thinking in all possible combinations, as in the question above?

A:

$1$ is ok.
So is $2a$, though you could have done less work by just multiplying one of the expressions by $\binom31$, instead of listing the $3$ combinations.
For $2(b)$, there is a beautiful way to avoid all computations !
By symmetry, P(last ball is white) = P(first ball is white) = $\dfrac37$
PS:
What I am saying is that since balls don't have preference for positions, the probability of a white ball being found in any position = $\frac37$, the same as that for the first position.
Let's check for position $2$ as an example.
P(white ball found in position $2$) =$P(BW) + P(WW) = \frac47\cdot\frac36 + \frac37\cdot\frac26 = \frac 37$

