# Thread: In a family of 5 children, determine the following probabibilities?

1. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
First off let's make sure you understand I"m not trying to argue with you for the sake of arguing with you. I'd like to understand this.

For (a) and (b) I agree with you. But (c) now implies an ordering on the children by age. Thus they do have at least the further property of age. If I can order them by age I can individually tag them, unless two happen to be born at precisely the same time in which case the ordering isn't well defined.

If I can individually tag them then each of them is distinguishable via their tag.

What am I getting wrong?
Let's forget probabilities for a moment and start instead by listing off all of the possible ways to have five children of various genders. The children are ordered from oldest to youngest. We use a sequence of B's and G's to represent the sequence of genders of the children. We know that there will be 32 different sequences. Part (c) is asking among those 32 sequences, how many of them will begin with 3 B's and end with 2 G's? As Plato said, there is exactly one such sequence. The order of the ages tells you the order that the letters must be B or G, but no more than that. It does not allow you to distinguish further between two sequences.

2. ## Re: In a family of 5 children, determine the following probabibilities?

Yes I understand that. But again you are referring to all the boys as b or 1 or whatever and all the girls as g or 0.

My contention is that given an ordering you can label the boys b1 b2 b3 and the girls g1 g2 and when this is done the sequence

b1 b2 b3 g1 g2 is not the same as the sequence b2 b1 b3 g1 g2

I'm going to leave it at this point. If you all say the question implies the children are indistinguishable other than gender I'll take your word for it.

3. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
If you all say the question implies the children are indistinguishable other than gender I'll take your word for it.
No one said that the "children are indistinguishable", at least I certainly did not. Of course they are.
I just say that it has nothing to do with the question.
I will say it again: The question is not about the children, it is about the gender of the children.

If I flip a coin five times, what is the probability of $HHHTT~?$
Maybe you can convince us that is different from part c of the OP?

P.S.
Suppose that we have a collection of five coins: a penny, a nickel, a dine, a quarter, and a half-dollar. Flip each coin in some random order. Record the outcomes as $H\text{ or }T$.
What is the probability of $HHHTT~?$

4. ## Re: In a family of 5 children, determine the following probabibilities?

Originally Posted by romsek
Yes I understand that. But again you are referring to all the boys as b or 1 or whatever and all the girls as g or 0.

My contention is that given an ordering you can label the boys b1 b2 b3 and the girls g1 g2 and when this is done the sequence

b1 b2 b3 g1 g2 is not the same as the sequence b2 b1 b3 g1 g2

I'm going to leave it at this point. If you all say the question implies the children are indistinguishable other than gender I'll take your word for it.
My point is, you are labeling the boys and the girls. Then, you are choosing which of the children is a specific boy or girl based on age. So, if that is the case, then you are using a different system of possible outcomes. There are more than 32 total possible outcomes.

Suppose there are zero boys and five girls. Then there are 5! different orders for the five girls. Suppose there is one boy and four girls. Then there are 5! different orders for those labels. Hence, there are $32\cdot 5!$ different possible outcomes. Now, among those outcomes, you are saying that there are $\binom{5}{3}\cdot 3!\cdot 2!$ different outcomes where the three oldest are boys and the two youngest are girls. Hence the probability is:

$\frac{\binom{5}{3}\cdot 3!\cdot 2!}{32\cdot 5!} = \frac{1}{32}$

This gives the same answer as what Plato got.

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