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?
If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.