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Boy or Girl: Two Interpretations


Date: 01/03/2000 at 20:09:56
From: Phil Reid
Subject: Changing Probability: Two Boys

While looking for something else, I stumbled upon this question:

  Probability of Two Male Children
  http://mathforum.org/dr.math/problems/mcclory.7.5.96.html   

and saw that you said the probability of the second boy having a 
brother was 1/3. By my calculations the boy's sibling is either elder 
or younger and either male or female. Assuming that the probabilities 
of each are equal:

     elder brother   = 1/4
     younger brother = 1/4
     elder sister    = 1/4
     younger sister  = 1/4

As such, the probability of him having a brother is 1/4 + 1/4 = 1/2.
 
Where you went wrong was in saying b:b was as likely as b:g or g:b, 
when of course in b:b if we now call the boy we know about x, we have 
x:b and b:x, so we have the sets x:g, g:x, b:x, and x:b; and thus a 
probability of 1/2.


Date: 01/04/2000 at 07:39:56
From: Doctor Anthony
Subject: Re: Changing Probability: Two Boys

Below is a similar question from another enquirer and my reply.

   This is an old question in my high school textbook but I can't 
   accept the answer:
 
   Mr. Smith has two children. You already know that one of them is a 
   girl; what's the probability that Mr. Smith has one boy and one 
   girl? 
  
   Official Solution: 2/3

   By listing [BB,BG,GB,GG], we know now one of them is a girl, so 
   cancel BB, therefore probability is 2/3. 

   I can't believe this because I think whether the other child is a 
   boy or girl is independent of this child. So the probability should 
   be 1/2.

Probability depends on the way you define the sample space. It is easy 
to obtain different answers if you change the sample space. A well- 
known example is the probability of drawing a random chord in a circle 
such that its length is greater than the side of the inscribed 
equilateral triangle. The probability can be 1/4, 1/3 or 1/2 depending 
on how you define a 'random' chord. None of these answers is more 
'correct' than the others; they are all correct if the randomness is 
defined in a particular way.

Now going to your problem with the two children. If the sample space 
is the total ways that you could have two children then BB, BG, GB, GG 
is correct and if you cancel BB then probability of one boy, one girl 
is 2/3. However, if you take the sample space as having one child, 
then B or G are the two possibilities, and the probability of a boy is 
1/2. So if you ask the question BEFORE the second birth the 
probability of a boy is 1/2. If you ask the question AFTER both 
children are born then the probability is 2/3. In other words, 
probability is not a fixed property of some event, it is a function 
RELATIVE to a framework of events.
 
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/08/2000 at 13:47:24
From: John Paul
Subject: Re: Changing Probability: Two Boys

Sorry, but you are refusing to accept that in the sample which you 
describe as b:b there is both b1:b2 and b2:b1. The probability of an 
event with two independent equal choices is always 1/2 for either 
one...


Date: 02/07/2000 at 12:52:24
From: Doctor TWE
Subject: Re: Changing Probability: Two Boys

Hi and thanks for writing to Dr. Math.

One of the biggest problems in probability is stating the problem 
clearly. Either answer, your 1/2 or Dr. Anthony's 2/3, could be 
correct depending on how the problem is set up.

In this problem, a key factor in determining the probability is how 
the child and family are selected. When we say, "in a two-child 
family, one child is a boy," how did we select the child? The 
selection process makes a big difference in the final probability (or, 
as Dr. Anthony would say, in the "sample space" of the problem.)
     ________________________________

Supposing that we randomly pick a _child_ from a two-child family. We 
see that he is a boy, and want to find out whether his sibling is a 
brother or a sister. (For example, from all the children of two-child 
families, we select a child at random who happens to be a boy.) In 
this case, an unambiguous statement of the question could be:

     From the set of all families with two children, a child is 
     selected at random and is found to be a boy. What is the 
     probability that the other child of the family is a girl?

Note that here we have a pool of kids (all of whom are from two-child 
families) and we're pulling one kid out of the pool. This is like the 
problem you're talking about. The child selected could have an older 
brother, an older sister, a younger brother or a younger sister.

Let's look at the possible combinations of two children. We'll use B 
for Boy and G for girl, and for each combination we'll list the older 
child first, so GB means older sister while BG means younger sister. 
There are 4 possible combinations:

     {BB, BG, GB, GG}

From these possible combinations, we can eliminate the GG combination 
since we know that one child is a boy. The three remaining possible 
combinations are:

     {BB, BG, GB}

In these combinations there are four boys, of whom we have chosen one. 
Let's identify them from left to right as B1, B2, B3 and B4. So we 
have:

     {B1B2, B3G, GB4}

Of these four boys, only B3 and B4 have a sister, so our chance of 
randomly picking one of these boys is 2 in 4, and the probability is 
1/2 - as you have indicated.
     ________________________________

But now let's look at a different way of selecting the "boy" in the 
problem. Suppose we randomly choose the two-child _family_ first. Once 
the family has been selected, we determine that at least one child is 
a boy. (For example, from all the mothers with two children, we select 
one and ask her whether she has at least one son.)  In this case, an 
unambiguous statement of the question could be:

     From the set of all families with two children, a family is 
     selected at random and is found to have a boy. What is the 
     probability that the other child of the family is a girl?

Note that here we have a pool of families (all of whom are two-child 
families) and we're pulling one family out of the pool. Once we've 
selected the family, we determine that there is, in fact, at least one 
boy.

Since we're told that one child (we don't know which) is a boy, we can 
eliminate the GG combination. Thus, our remaining possible 
combinations are:

     {BB, BG, GB}

Each of these combinations is still equally likely because we picked 
one of the four families.

Now we want to count the combinations in which the "other" child is a 
girl. There are two such combinations: BG and GB.

Since there are three combinations of possible families, and in two of 
them one child is a girl, the probability is 2/3.
     ________________________________

Why are these two probabilities different?

As in other probability problems, how information is obtained is as 
important as the information itself. Without knowledge of the data 
gathering process, ambiguity can result. How do we know that one child 
is a boy?

In the first (your) interpretation, each _boy_ has an equal chance of 
being chosen. Thus, the family with two boys has twice the chance of 
being the "chosen family." The boys are equally probable, but the 
families are not.

In the second (Dr. Anthony's) interpretation, each _family_ has an 
equal chance of being chosen. In a family with two boys, each boy has 
only half that chance of being "the boy" referenced in the statement. 
The families are equally probable, but the boys are not. In this case, 
the two "events" are not independent, because we're selecting a 
family, not an individual child. In fact, there's really only one 
"event" - the selection of the family.

If you're still not convinced, try the following experiment. Take two 
fair coins and toss them. I think you'll agree that each coin has a 
1/2 chance of being heads. On each toss, see if at least one of the 
coins is heads (the equivalent of "at least one child is a boy"). If 
both coins are tails (both children are girls), ignore the outcome and 
toss again. If at least one of the coins is heads, record whether you 
had two heads (the boy has a brother) or a head and a tail (the boy 
had a sister). Over many tosses, you should find yourself getting 
about twice as many head-tail tosses as head-head tosses. Of course, 
if you count each head-head toss twice (once for each head tossed)...

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/   



Date: 02/27/2000 at 07:46:43
From: John Paul
Subject: Re: Changing Probability: Two Boys

Okay, I'm not completely happy with this, but unless I can be bothered 
to go find a math student and get him to look at this, I can't really 
say much more.

Phil


Date: 02/28/2000 at 12:30:11
From: Doctor Twe
Subject: Re: Changing Probability: Two Boys

Hi again - thanks for writing back.

Many first-year probability textbooks address this question 
specifically. My college probability book was _Fundamentals of 
Probability_ by Saeed Ghahramani of Towson State University (Prentice 
Hall, 1996). That book addresses this problem in examples 3.2 and 3.3 
on pp. 73-74. You may want to look there.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Probability
High School Probability

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