Conditional Probability Question EDIT!!!

a)  ￼The probability that any child in a certain family will have blue eyes is 1/4, and this feature is inherited independently by different children in the family.  If there are five children in the family and it is known that at least one of these children has blue eyes, what is the probability that at least three of the children have blue eyes?
b)  Consider a family with the five children just described.  If it is known that the youngest child in the family has blue eyes, what is the probability that at least three of the children have blue eyes?

Hello!  I'm pretty sure I understand part (a), but I'm not sure about part b.  Since the child is actually distinguished in this case, does it change the denominator?
So since its distinguished who actually has the blue eyes, namely, the young child, does the denominator just become 1/4?
Actually it turns out that the denominator is actually 1.  Can someone please explain?
Please note that the answers in (a) and (b) are actually different.  They are NOT the same.  It turns out that knowing who the child is actually simplifies the problem and removes the necessity of using conditionals.
Thanks in advance!
Part a:
Let:
A= event that at least 3 children have blue eyes
B= event that at least 1 child has blue eyes
$\therefore A \subset B$
$\Pr(A \mid B)=\cfrac{\Pr(A \cap B)}{\Pr(B)}=\cfrac{\Pr(A)}{\Pr(B)}=\cfrac{ \sum\limits_{i=3}^5 \binom{5}{i} \cdot (0.25)^i \cdot 0.75^{5-i}}{1-0.75^5}=0.1357\tag{1}$
Part b:
A=event youngest child has blue eyes
B=event at least 3 children have blue eyes
$\Pr(A \mid B)=\cfrac{ \sum\limits_{i=2}^4 \binom{4}{i} \cdot (0.25)^i \cdot 0.75^{4-i}}{1/4??}\tag{2}$

I know that a lot of time has passed, but FWIW here's my take.

b) Consider a family with the five children just described. If it is
  known that the youngest child in the family has blue eyes, what is the
  probability that at least three of the children have blue eyes?

If "it is known" implies truth, i.e. P(youngest child has blue eyes) = 1.
Except that now, your event $A$ is no longer "At least three children have blue eyes." It is now at least two children have blue eyes.
Now you can sum across your probabilities as usual:
$$
p(A|B) = \dfrac{\sum^4_{i=2}\binom{4}{i}\frac{1}{4}^i\frac{3}{4}^{4-i}}{1}
$$
Another way of thinking about the problem is that you define a new scenario. You have a group of 4 children, probability for having blue eyes as $p_2$, and the problem is to find the probability that at least 2 have blue eyes. There is no longer any conditional probability, but if you find $p_2$, you get the answer to the question posed at the beginning.
