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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

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

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  • $\begingroup$ In the second calculation, I think you've miscalculated $\Pr(A \cap B)$. You can rewrite $A \cap B$ as $A' \cap B$, where $A'$ is the event that at least two of the four older children have blue eyes. By independence, $\Pr(A' \cap B) = \Pr(A') \Pr(B)$, and the thing in your numerator is just $\Pr(A')$. The $\Pr(B)$ you left out of the numerator will cancel the $\Pr(B)$ in the denominator when you put it back in. $\endgroup$ – Vectornaut Sep 24 '13 at 15:50
  • $\begingroup$ The mechanical approach that you're using for this problem works perfectly well---as it must, if the rules of probability are consistent. However, there should be a more enlightening way of solving the problem, based on user96614's insight that knowing the youngest child has blue eyes "gives you the same information" as knowing at least one child has blue eyes. I'll try to write an answer about that later. $\endgroup$ – Vectornaut Sep 24 '13 at 15:53
  • $\begingroup$ Now I've done both calculations by hand for the case where there are three children, and you're trying to decide how likely it is that at least two have blue eyes. Unless I've made a mistake, the "youngest child has blue eyes" condition and the "at least one child has blue eyes" conditions give different conditional probabilities, so I have to retract what I said before about the two conditions "giving you the same information." I'll try to understand why this is... $\endgroup$ – Vectornaut Sep 24 '13 at 16:25
  • $\begingroup$ For the second problem, suppose you want the denominator to be $1/4$ (which sounds reasonable). Then for the numerator you want the probability the youngest is blue and there are at least $3$ blue. This is the probability youngest is blue times the conditional probability of at least $3$ blue given youngest is blue. The second term is just the probability of $\ge 2$ blue among the first four. The point is that if we use denominator $1/4$, then in our calculation we will need to use the conditional probability asked for. Then we multiply and divide by $1/4$. $\endgroup$ – André Nicolas Sep 24 '13 at 16:37
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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.

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  • $\begingroup$ No you are incorrect. The answer are not the same. $\endgroup$ – user1527227 Sep 24 '13 at 15:04
  • $\begingroup$ I'm pretty sure that the right is given by @ADB below $\endgroup$ – Dan Kruchinin Nov 22 '14 at 11:56
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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.

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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.

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  • $\begingroup$ "If, on the other hand, someone picks a coin without regard to whether it is a head or a tail" <-- I assume in this scenario it's also not known whether at least one coin is a head, in contrast to the scenario in the preceding paragraph? $\endgroup$ – Joseph Garvin May 13 '17 at 20:44
  • $\begingroup$ Also why in the second scenario is the sample space narrowed further? It sounds like they are introducing the information that at least one coin is a head (unless you mean they always pick the first coin?) but your narrowed sample space indicates afterwards you know the head is the first coin specifically. Since we're distinguishing HT and TH, if they don't always pick the first coin, do we at least see which coin they pick? $\endgroup$ – Joseph Garvin May 13 '17 at 20:50
  • $\begingroup$ Ugh, by second scenario I meant last scenario, but stack exchange doesn't let you edit comments after 5 minutes. $\endgroup$ – Joseph Garvin May 13 '17 at 20:56
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Under you notations, the answer of the second part should be rather $$ Pr(B|\,A)=\frac{Pr(A\cap B)}{Pr(A)}=\frac{Pr(C)}{\frac{1}{4}} $$ where $C=$ at least 2 children among the 4 children left have blue eyes and "the youngest child has blue eye" .

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  • $\begingroup$ denominator is actually just 1 $\endgroup$ – user1527227 Sep 24 '13 at 15:43
  • $\begingroup$ sorry for my mistake, actuallely $C$ represents the event "at least 2 children among the 4 children left have blue eyes and the youngest child has blue eye" Hence you should add $\frac{1}{4} \times$ in your numerator which compensate the denominator $Pr(A)=\frac{1}{4}$ $\endgroup$ – CKH Sep 24 '13 at 15:51

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