Q.A couple has two children,

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

(i) Sample space for two children: $\{ MM, MF, FM, FF \}$.

Given at least one male, the reduced space is $\{ MM, MF, FM \}$.

Only $MM$ satisfies "both males".

So $P = \frac{1}{3}$.

(ii) Given the elder child is female, the reduced space is $\{ FM, FF \}$.

Only $FF$ satisfies "both females".

So $P = \frac{1}{2}$.

Concept and Intuition

When we say "given that" something is true, we are no longer looking at all possible outcomes — we restrict our attention to only those outcomes where the condition holds. This is the heart of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, where $B$ is the condition.

For a family with two children, the natural sample space (assuming equal probability for male and female, and independence) is:

Each outcome has probability $\frac{1}{4}$. The order matters here — first child then second — so MF and FM are distinct.

The classic mistake is to treat "at least one male" as if it only eliminates FF, but then to forget that the remaining three outcomes are not equally likely under the condition? Actually, they are equally likely because each original outcome had equal probability, and we are simply discarding one. So the conditional probability is just counting: number of favorable outcomes in the reduced space divided by total outcomes in the reduced space.

Let's work each part carefully.

(i) Both males, given at least one male

1. Define events.

   Let $A$ = "both children are males" = $\{MM\}$.

   Let $B$ = "at least one child is male" = $\{MM, MF, FM\}$.

2. Find the reduced sample space.

   The condition $B$ removes only $\{FF\}$. So the new sample space has 3 equally likely outcomes: MM, MF, FM.

3. Count favorable outcomes.

   Only MM satisfies $A$. So exactly 1 outcome.

4. Compute probability.

(ii) Both females, given elder child is female

1. Define events.

   Let $C$ = "both children are females" = $\{FF\}$.

   Let $D$ = "the elder child is female" = $\{FF, FM\}$.

2. Find the reduced sample space.

   Condition $D$ keeps only outcomes where the first child (elder) is female: FF and FM. These are equally likely.

3. Count favorable outcomes.

   Only FF satisfies $C$. So 1 outcome.

4. Compute probability.

Method: Conditional probability by reducing the sample space

For "given that … , find the probability that …" problems with a small set of equally likely outcomes, you can skip the fraction formula and just shrink the world of possibilities.

Steps

Step 1: Write the full, ordered, equally likely sample space.

List every outcome once, keeping order where it matters (elder child first, then younger): $\{MM, MF, FM, FF\}$, each equally likely.

Step 2: Keep only the outcomes that satisfy the given condition.

The condition becomes your new, smaller sample space. "At least one male" keeps $\{MM, MF, FM\}$; "elder is female" keeps $\{FF, FM\}$. Discard everything the condition rules out.

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

The whole answer turns on listing the reduced space completely — order-sensitive outcomes like $FM$ versus $MF$ are the ones most often missed.

Common Mistakes

Mistake 1: Reducing "at least one male" to $\{MM, MF\}$ and getting $\tfrac12$.

Why it's wrong: it drops $FM$ (elder female, younger male), which also has at least one male. Correct approach: the reduced space is $\{MM, MF, FM\}$, so the answer is $\tfrac13$.

Mistake 2: Treating "at least one male" as "the elder child is male".

Why it's wrong: those are different conditions that shrink the sample space by different amounts. Correct approach: keep every outcome with a male anywhere, not just elder-male ones.

Mistake 3: In part (ii), forgetting that "elder is female" fixes only the first child.

Why it's wrong: the reduced space is $\{FF, FM\}$, not a single outcome. Correct approach: count $FF$ among these two, giving $\tfrac12$.