Classic Problems || Dr. Math FAQ || About Dr. Math || Search Dr. Math || Dr. Math Home

For a review of basic concepts, see Introduction to Probability and Permutations and Combinations.
When the only information given is that there are two children and one is a boy, here are two ways of looking at the problem: ## Combinations in the Sample Space-
In a two-child family, there are four and only four possible combinations of
children. We will label boys B and girls G; in each case the first letter
represents the oldest child:
- {BB, BG, GB, GG}
When we know that one child is a boy, there cannot be two girls, so the sample space shrinks to:
- {BB, BG, GB}
Two of the possibilities in this new sample space include girls:
- {BG, GB}
and since there are two combinations out of three that include girls, the probability that the second child is a girl is 2/3. ## Probability Tree-
We can also visualize this problem using a probability tree.
In the following probability tree, the number next to each letter (B for boy, G for girl) indicates the probability of that event. To calculate the unconditional probability of any one combination of children, we multiply the numbers along that combination's path. Given no information other than that the family has two children, the four combinations of boys and girls are equally likely to occur. Here's an unconditional probability tree for a two-child family: First Second Unconditional Child Child Probability __B(1/2)__ 1/4 | __B(1/2)__| | |__G(1/2)__ 1/4 | Family -| | __B(1/2)__ 1/4 |__G(1/2)__| | |__G(1/2)__ 1/4To turn an unconditional probability tree into a conditional probability
tree, we delete the paths that are not relevant to the new conditions given with
the problem. The unconditional probability for each combination remains the
same, but the conditional probability is calculated from the new revised sample
space.
First Second Unconditional Conditional Child Child Probability Probability __B(1/2)__ 1/4 1/3 | __B(1/2)__| | |__G(1/2)__ 1/4 1/3 | Family -| | __B(1/2)__ 1/4 1/3 |__G(1/2)__|Since each of the remaining three combinations of children is equally likely, the conditional probability of each combination is 1/3. The conditional probability creates a new revised sample space, so the probability that a two-child family will include one girl is the sum of the probabilities of
the combinations that include a girl.Since two of the three members of the sample space include a girl, the probability that the second child is a girl is 1/3 + 1/3 = 2/3. Many people imagine that this probability must be wrong since they think girls should be as likely as boys, but given the information that one child is a boy, a 2/3 probability that the other child is a girl is correct.
Suppose we're told that the oldest child is a boy? Let's look at this new
information using the same two methods:
## Combinations in the Sample Space-
When the oldest child is a boy, the original sample space:
- {BB, BG, GB, GG}
shrinks to: - {BB, BG}
because the two other possibilities, {GB, GG}, show girls as the oldest child and thus are no longer possible. Since only one of the possibilities in the new sample space, {BG}, includes a girl, the probability that the second child in the family is a girl is 1/2. ## Probability Tree-
For this variation of the problem we construct the same conditional probability
tree as above, but since in our notation the first letter represents the oldest
child, we delete the bottom two paths where both combinations represent an
oldest daughter:
First Second Unconditional Conditional Child Child Probability Probability __B(1/2)__ 1/4 1/2 | __B(1/2)__| | |__G(1/2)__ 1/4 1/2 | Family -|To find the probability that the younger child is a girl, follow the second path. The probability is 1/2.
Remember: information that creates conditional probability can
dramatically affect common sense ideas about probability. For example, no matter
how unlikely it may seem to you, if you meet a mother of two who says she has a
daughter, a basic knowledge of probability tells you there's a 2/3 probability
that the daughter mentioned has a brother. If she says she's an older
daughter, you know there's a 1/2 probability that the daughter has a younger
brother.
## From the Dr. Math archives: |

[**Privacy Policy**]
[**Terms of Use**]

Math Forum Home ||
Math Library ||
Quick Reference ||
Math Forum Search

© 1994-2015 Drexel University. All rights reserved.

http://mathforum.org/

The Math Forum is a research and educational enterprise of the Drexel University School of Education.