Conditional probability of at least two girls with eldest daughter

Key takeaway: The event “eldest child is a girl” changes the sample space for this family probability question.

Build the correct sample space

Because the eldest child is already known to be a girl, the first child is fixed as G. That means only the remaining three children are uncertain. Each of those three can be a girl or a boy, so the reduced sample space has $2^3 = 8$ equally likely outcomes.

We can list them as G followed by the outcomes for the other three children: GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB.

Count the favorable outcomes

The question asks for the probability that there are at least two girls in total. Since the eldest is already a girl, we need at least one more girl among the remaining three children.

The easiest way is to count the complement. The only way to have fewer than two girls overall is to have no additional girls among the last three children, which means all three are boys: GBBB. That is 1 outcome out of 8.

So the favorable probability is

$1 - \frac{1}{8} = \frac{7}{8}.$

That matches option D.

Why the conditional information matters

The phrase “if the eldest child is a girl” is not extra decoration. It changes the probability model. Without that condition, a family of four children would have $2^4 = 16$ equally likely gender patterns, and the answer would be different. Conditioning reduces the space before counting.

A helpful way to think about it is that the eldest child is no longer random in this question. You are only deciding the genders of the three younger children. Once that is clear, the problem becomes a short counting exercise rather than a full four-child probability calculation.

Common wrong approach

Some students count all four children and then simply divide by 16, forgetting that the eldest has already been fixed as a girl. That leads to overcounting and often produces 3/4, which is not correct here. Others count only the cases with exactly two girls instead of at least two girls. In this setup, outcomes with three or four girls also count as favorable.

Final answer

The probability is 7/8, so the correct choice is D.

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Pitfalls the pros know 👇 A frequent trap in this family probability problem is to ignore the conditional phrase about the eldest child. Once the eldest is known to be a girl, the experiment no longer has 16 equally likely outcomes; it has only 8. Another error is to treat “at least two girls” as “exactly two girls.” That would omit the cases with three girls or four girls, both of which are favorable. The cleanest check is to work from the complement: with the eldest fixed as G, the only bad outcome is the other three children all being boys. That gives one bad case out of eight, so the answer must be 7/8. If your count does not match that complement logic, the sample space has probably been built incorrectly.

What if the problem changes? If the question changed to a family of four children where the eldest child is a boy, and you were asked for the probability of at least two girls, the same method would still work. The eldest would be fixed as B, leaving three random children. The favorable cases would now require at least two girls among those three, which can be counted as exactly two girls or three girls. There are 3 outcomes with exactly two girls and 1 outcome with three girls, for 4 favorable outcomes out of 8 total, giving 1/2. If the condition were removed entirely, the sample space would return to 16 outcomes, and a different answer would be needed. This shows how conditioning changes the denominator before any counting begins.

Tags: conditional probability, equally likely outcomes, complement counting