Q.A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

The key idea is conditional probability — we restrict the sample space to only those outcomes where the condition holds.

Step 1: List all equally likely outcomes for two children (order matters: older, younger):

$\{ BB, BG, GB, GG \}$.

Step 2: The condition “at least one is a boy” removes $GG$, leaving the reduced sample space:

$\{ BB, BG, GB \}$.

Step 3: Among these three outcomes, only $BB$ has both children as boys.

So the required probability is $\frac{1}{3}$.

Why conditional probability?

When we say "given that at least one is a boy," we are no longer considering all possible families with two children. We are conditioning on a specific event — we only look at families that satisfy the condition. This shrinks the sample space. The probability we want is the fraction of those families where both children are boys.

A common mistake is to think: "If one is a boy, the other is either a boy or a girl, so it's 1/2." That reasoning is wrong because it treats the children as unlabeled. In reality, the two children are distinct individuals (say, older and younger), and the condition "at least one boy" includes three distinct cases, not two.

Step-by-step solution

1. List the sample space for two children. Each child can be a boy (B) or a girl (G). Assuming equal probability and independence, the four equally likely outcomes are:

Here, the first letter denotes the older child, the second the younger. Each outcome has probability $\frac{1}{4}$.

2. Identify the conditioning event. The condition is "at least one is a boy." This event, call it $A$, includes all outcomes except $GG$:

So $P(A) = \frac{3}{4}$.

3. Identify the event of interest. We want "both are boys," call it $B$:

So $P(B) = \frac{1}{4}$.

4. Apply the conditional probability formula. The probability of $B$ given $A$ is:

Since $B$ is a subset of $A$ (if both are boys, then certainly at least one is a boy), we have $B \cap A = B$. Thus:

Method: Conditional probability by shrinking the sample space

Use this whenever a condition ("given that …") can be described as a set of equally likely outcomes — you can then count instead of using the ratio formula.

Steps

Step 1: Write the full equally-likely sample space, keeping items distinguishable.

List every outcome so that all are equally likely. When objects look identical (two children, two dice), label them by order (older/younger, first/second) so cases like $BG$ and $GB$ stay distinct — collapsing them is the classic error.

Step 2: Discard every outcome where the condition fails.

Keep only outcomes consistent with the "given" event. This reduced set is your new universe.

Step 3: Count favourable outcomes inside the reduced space and divide.

This equals the formula $P(A\cap B)/P(B)$ but is faster and less error-prone when outcomes are equally likely.

Common Mistakes

Mistake 1: Answering $\tfrac12$ by treating the "other" child as a fresh coin flip.

Why it's wrong: "at least one boy" is not the same as "the first child is a boy"; it keeps three equally likely families $BB, BG, GB$, not two. Correct approach: condition on the three-outcome reduced space, where only $BB$ works, giving $\tfrac13$.

Mistake 2: Merging $BG$ and $GB$ into a single case.

Why it's wrong: the children are distinguishable by birth order, so $BG$ and $GB$ are separate equally likely outcomes. Correct approach: keep ordered outcomes when listing the sample space so the counts stay correct.