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
$\Pr(A \mid B)=\cfrac{ \sum\limits_{i=2}^4 \binom{4}{i} \cdot (0.25)^i \cdot 0.75^{4-i}}{1/4??}\tag{2}$
Best Answer
This is a question from the book 'All of Statistics' by Larry Wasserman.
A simple example to illustrate the concept...
Say you record the outcome of the toss of two fair coins. The sample space is {HH, HT, TH, TT}.
If it is known that the first coin is a head, then the probability of getting two heads will be 1/2. The sample space is reduced to {HT, HH} and only {HH} gives the desired outcome.
If it is known that as least one of the coins is a head, then the probability of getting two heads is 1/3. The sample space is reduced to {HH, HT, TH} and only {HH} gives the desired outcome.
Now, say it is known that at least one of the coins is a head. If someone else looks at the coins, picks out a head coin because it is a head, and tells you 'this coin is a head', the probability of two heads is still 1/3.
If, on the other hand, someone picks a coin without regard to whether it is a head or a tail, and tells you 'this coin is a head', then the sample space is reduced to {HH, HT} and the probability of two heads is 1/2.
The answer to the original question hinges on whether the child was chosen because she was the youngest child, or because she had blue eyes.
Apologies for not writing as a comment, I do not have any rep.