Conditional probability question

[hholzer]

Consider this scenario:

"From families with three children, a family is selected at random and found
to have a boy. What is the probability that the boy has an older brother
and a younger sister? Assume that in a three-child family all gender
distributions have equal probabilities."

So we operate in the space of:
(given a triplet (xyz) take it to mean x > y > z by age;
to distinguish boys, the chosen boy will be denoted by
*)

S* = { (bbb*) , (bb*b) , (b*bb) , (b*bg) , (bb*g) , (b*gb) , (bgb*) ,
(gb*b) , (gbb*) , (gb*g) , (ggb*) , (b*gg) }

then, how is P( {ggb*} ) = P( {gb*g } ) = P( {b*gg} ) = 1/7

I'm not seeing why we assign 1/7 to this event though.

Elaboration welcomed.

 

[mathman]

There are seven (equally likely) possibilities for a family to have three children and at least one boy: bbb, bbg, bgb, gbb, bgg, gbg, ggb.
The second case (bbg) is the only one in which the answer may be yes. In that case the probability is 1/2, making the overall probability 1/14.

 

[SW VandeCarr]

There are 7 ways for a family of 3 to have at least one boy. Of these, there are 4 ways that the middle child is a boy: bbg, gbg, gbb, bbb. So the probability of bbg is (4/7)(1/4)=1/7. Or you could simply say the probability of bbg is 1/7 given all the possibilities of at least one boy.

EDIT: We can get seven ways with at least one boy by realizing there are $$2^3$$ combinations of three independent binary variables and ggg is the only combination that excludes boys.

 

[hholzer]

It looks like you two reached different answers. I know that the event
the middle child is a boy with an older brother and younger sister
is 1/14.

I was asking about why:

P( {ggb*} ) = P( {gb*g } ) = P( {b*gg} ) = 1/7

With a reduced sample space of:

(given a triplet (xyz) take it to mean x > y > z by age;
to distinguish boys, the chosen boy will be denoted by
*)

S* = { (bbb*) , (bb*b) , (b*bb) , (b*bg) , (bb*g) , (b*gb) , (bgb*) ,
(gb*b) , (gbb*) , (gb*g) , (ggb*) , (b*gg) }

Could you elaborate with:
1) if you use a definition, specify it
2) if you create an event, define it
3) if a theorem is invoked, specify it

Thanks in advance.

 

[statdad]

Why not use the definition of conditional probability.
If A is the event you're interested in, and B is the event that a family has three children,

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\dfrac{1}{8}}{\dfrac{7}{8}}=\frac 1 7$$

 

[hholzer]

I was puzzling over my book's use of a reduced sample space
with the format of a triplet listed above. Their sample space
had 12 points in it and so I was not arrive at a probability of
1/7.

Where the reduction in sample space is Q(A) = P(A|B)
B our sample.

 

[SW VandeCarr]

It looks like you two reached different answers. I know that the event the middle child is a boy with an older brother and younger sister is 1/14.

How do you know this?

You're making this way too complicated. You can approach this problem several ways. Statdad's and mine are two ways. The ordered triplet bbg represents one of seven possibilities with at least one boy. It represents the birth order: boy(older brother), boy (subject), (girl) younger sister.

 

[hholzer]

@VandeCarr: My comment was addressed to your and mathman's solution. Not statdad's. statdad responded after you presented a solution and my response preceded statdad's. So it isn't possible that I was referring to your's and statdad's.

I know the probability of the event being a boy with an older brother and younger sister
is 1/14 because I have the solution manual. I was trying to understand the presented solution to the problem, but it was not making sense. Too, to put this into context: I'm specifically trying to understand the problem in terms of a reduced probability space. My book defines that as follows:

Let B be an event of a sample space S with P(B) > 0. For a subset A of B, define
Q(A) = P(A|B). Then Q is a function from the set of subsets of B to [0,1].

Here is the solution:
http://i53.tinypic.com/v5l5pi.png

Which the problem statement mentioned in the original post corresponds to
(a). They comment "the reduced sample space is:"

There are a couple of things I do not see here: if we use a reduced
sample space, redefining a probability function Q in terms of conditional
probability with probability function P, then what was the original sample
space? If the reduced one has 12 points, then what did the original one
have? That is one thing that is unclear to me.

The other thing that is unclear is what I had mentioned previously:
the events {b*gg}, {gb*g}, {ggb*} correspond to a probability of
1/7. You'll note the solution uses the function P (not a function Q)
but claims to use a "reduced sample space." That bothers me because
Q should be used with consistency with respect to a
reduced sample space and P for the complete sample space(i.e.,
no conflating of Q and P). As stated by statdad, his solution is clear
and agrees with the use of the function P; which isn't the case for
the book's approach as previously mentioned.

 

[SW VandeCarr]

Consider this scenario:

"From families with three children, a family is selected at random and found
to have a boy. What is the probability that the boy has an older brother
and a younger sister? Assume that in a three-child family all gender
distributions have equal probabilities."
Elaboration welcomed.

On reconsideration, the probability of 1/14 is correct if you designate a particular boy in advance. The question above can be interpreted as the family selected "has a boy" but could have other boys. In fact, to satisfy the requirement of birth order bbg, it obviously must have exactly two boys. If saying "a boy" designates that boy's probability, then there are two possibilities: b*bg or bb*g. Since the probability of bbg is 1/7 in families with at least one boy, then the probability of bb*g is (1/2)(1/7)= 1/14. However, should the calculation really be conditional on the family having at least two boys, or exactly two boys instead? The issue is not with math, but with the definition of the probability space.