Girl Child over Boy Child

226.6K Views

In the Mexico City area, there are two Houses H1 and H2. Both H1 and H2 have two children each.
In House H1, The boy plays for Mexico Youth academy and the other child plays baseball.
In House H2, The boy Plays soccer for his school in Mexico and they recently have a newborn.

Can you prove that the probability of House-H1 having a girl child is more than that of House-H2?

Share
Add Comment

  • 1 Answer(s)

    Let’s consider the scenario where the second child in H1 is a girl. In this case, we know that H1 has two children: one boy who plays for the Mexico Youth academy and one girl who plays baseball. The probability of H1 having a girl child is  1/2, because one of the two children is a girl.

    For H2, we know that they have two children: one boy who plays soccer for his school and a newborn, whose gender we do not know. there are two possible cases:

    • Case 1: H2’s newborn is a boy. In this case, H2 has two boys and no girl.
    • Case 2: H2’s newborn is a girl. In this case, H2 has one boy and one girl.

    Assuming that the probability of having a boy or girl is the same, the probability of Case 1 is 1/2, and the probability of Case 2 is also 1/2. Therefore, we can compute the overall probability of having a girl child in H2 as before:

    P(girl in H2) = P(girl in Case 1) * P(Case 1) + P(girl in Case 2) * P(Case 2) = 0 * 1/2 + (1/2) * 1/2 = 1/4

    So the probability of having a girl child in H2 is 1/4.

    Since the probability of having a girl child in H1 is now 1/2 and the probability of having a girl child in H2 is 1/4, we can conclude that the probability of House H1 having a girl child is actually higher than that of House H2 in this scenario.

    But if the 2nd child on house H1 is a boy then the answer is opposite.
    Moshe Expert Answered on 21st February 2023.
    Add Comment
  • Your Answer

    By posting your answer, you agree to the privacy policy and terms of service.
  • More puzzles to try-