Q.Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

Concept: Conditional Probability — we restrict the sample space based on the given condition.

Let the children be ordered (older, younger). The sample space for two children is:

$\{ BB, BG, GB, GG \}$, each equally likely.

1. Given the youngest is a girl Possible outcomes: $\{ BG, GG \}$ (only those where second child is a girl). Only $GG$ has both girls. Probability = $\frac{1}{2}$.
2. Given at least one is a girl …

Why Conditional Probability Works Here

When we say "given that" something is true, we are restricting our universe of possibilities. Instead of looking at all possible two-child families, we only look at those families that satisfy the condition. The probability then becomes the fraction of those restricted families that also have both girls.

The classic mistake is to treat the condition as if it were a separate event that doesn't change the sample space. But it does — and that's the whole point of conditional probability.

We'll use this formula, but more importantly, we'll build the restricted sample space by hand so the intuition is crystal clear.

Step-by-step solution

1. Write down the full sample space

For two children, each equally likely to be boy (B) or girl (G), the four equally likely outcomes are:

Each has probability $\frac{1}{4}$. The order matters: first child is older, second is younger.

2. Part (i): Given that the youngest is a girl

The youngest child is the second child in our ordered pairs. So the condition "youngest is a girl" means the second letter is G. The outcomes that satisfy this are:

Only these two families are possible now. Among them, how many have both girls? Only $GG$.

So the conditional probability is:

3. Part (ii): Given that at least one is a girl

Now the condition is broader: at least one of the two children is a girl. The outcomes that satisfy this are:

$\{ BG, GB, GG \}$ …

Method: Ordered sample spaces and the "at least one" trap

Use this for birth/gender-style problems (or any pair of equally-likely binary trials) where you condition on information like "the youngest is a girl" or "at least one is a girl".

Steps

Step 1: Use an ordered sample space.

List outcomes as ordered pairs so that position has meaning — for two children, $\{BB, BG, GB, GG\}$ with the first slot the elder and the second the younger, each of probability $\tfrac14$. Ordering is what lets "the youngest" pick out one specific slot.

Step 2: Turn the condition into a subset.

"Youngest is a girl" fixes the second slot: $\{BG, GG\}$. "At least one girl" removes only the all-boys outcome: $\{BG, GB, GG\}$. Notice these subsets have different sizes.

Step 3: Count the target within the conditioned subset.

$P(A\mid B)=\frac{n(A\cap B)}{n(B)}.$ …

Common Mistakes

Mistake 1: Answering $\frac12$ for part (ii).

Why it's wrong: "at least one is a girl" keeps three equally likely families $\{BG,GB,GG\}$, not two, and only $GG$ is favourable, so the answer is $\frac13$, not $\frac12$. Correct approach: build the restricted sample space explicitly instead of assuming the "other child" is a fresh 50-50.

Mistake 2: Treating parts (i) and (ii) as the same condition. …