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
$\Pr(A \mid B)=\cfrac{ \sum\limits_{i=2}^4 \binom{4}{i} \cdot (0.25)^i \cdot 0.75^{4-i}}{1/4??}\tag{2}$

The answer to part b should be same as that for part a. There is no more information in b as there is no information about correlation between age and eye color.
On further thought: the question is ambiguous because the answer depends on the experiment conducted because that influences the sample space for conditional probability.
Experiment 1: The researcher doesn't know anything any childs' eye color and then specifically checks the youngest child and finds that his/her eyes are blue. Then, this has not bearing on others and the answer to part b is just the unconditional probability that there are at least two children with blue eyes among the remaining four children.
Experiment 2: The researcher learns that at least one child has blue eyes. Then the researcher asks one of the possibly many children with blue eyes to step forward and then founds out that this child is the youngest (or the oldest or anything else). In this case, the answer to part b should be the same as the answer to part a.
