A woman has two children and it is known that at least one of the children is a boy. Under this information, what is the probability that both of her children are boys?

Introduction / Context:
This is a classic conditional probability problem related to the gender of children in a family. The key subtlety is the phrase at least one child is a boy, which changes the sample space compared to choosing two children without any extra information. Many students answer 1/2 by intuition, but that is not correct under the usual assumptions stated in such questions.

Given Data / Assumptions:

• The family has exactly two children.
• Each child is equally likely to be a boy or a girl.
• The gender of one child does not affect the gender of the other (independence).
• It is known that at least one of the children is a boy.
• We want the probability that both children are boys, given this information.

Concept / Approach:
We will list all equally likely possibilities for the two children and then condition on the information that at least one child is a boy. If we denote a boy by B and a girl by G, then ordered pairs (older, younger) cover all scenarios. We then remove any outcomes that contradict the given information and compute conditional probability: P(both boys | at least one boy) = (number of outcomes with both boys) / (number of outcomes with at least one boy).

Step-by-Step Solution:
Step 1: List all possible ordered pairs of genders for the two children: BB, BG, GB, GG. Step 2: Under the usual assumption, each of these four outcomes is equally likely. Step 3: The information at least one child is a boy rules out the outcome GG, where both are girls. Step 4: The reduced sample space is now {BB, BG, GB}. Step 5: Among these three remaining outcomes, only BB has both children boys. Step 6: Count favorable outcomes for both boys: there is 1 such outcome (BB). Step 7: Count all outcomes that satisfy at least one boy: there are 3 such outcomes (BB, BG, GB). Step 8: Conditional probability = favorable / total = 1 / 3.

Verification / Alternative check:
You can think of this in terms of long run frequencies. Imagine many families with two children, and every boy and girl arrangement is equally likely. Among 4 families on average, you expect 1 with GG, 1 with BB, and 2 with mixed genders BG and GB. If you now choose only those families that have at least one boy, you are choosing from 3 types: BB, BG, GB. Out of those 3 types, only 1 type, BB, has both boys. Therefore the probability is 1/3 rather than 1/2, confirming the conditional probability calculation.

Why Other Options Are Wrong:
Option 1/2 would be correct only if we started by knowing that exactly one child is a boy or if we looked at a completely different conditioning scenario. Option 2/3 would correspond to the probability that there is at least one boy, not that both are boys given that information. Option 1/4 is the unconditional probability of both children being boys with no extra information, which does not apply once we know at least one is a boy. Option 3/8 has no simple interpretation in this setup and does not come from any correct derivation.

Common Pitfalls:
Many students forget that conditional probability changes the sample space. They keep the denominator 4 and simply consider the 2 boys as one of two children, which is incorrect. Others assume that given at least one boy, the other child is equally likely to be a boy or girl, and directly answer 1/2, neglecting that the families with mixed genders appear twice in the original sample space (BG and GB). Always explicitly write the sample space and then cross out the outcomes that conflict with the given condition.

Final Answer:
The probability that both children are boys, given that at least one is a boy, is 1/3.