Explanation:
based on the rough assumption that the probabilty to get a boy is the same as to get a girl = 1/2 = 0.5, as there are only 2 possible outcomes.
when a family has 4 children, they have 2 possibilities for the first child, 2 for the second child, 2 for the third and 2 for the fourth child.
so, 2×2×2×2 = 2⁴ = 16 possibilities.
the probability for any of these 16 possibilities is
0.5⁴ = (1/2)⁴ = 1/16 = 0.0625
how many of these 16 possibilities include at least one boy ? all except the one, where all 4 children are girls.
so, 16 - 1 = 15
that means 15 out of possible 16 different "configurations" contain at least 1 boy.
the probabilty is therefore 15/16.
and that applies as mean value of 2000 families to
2000 × 15/16 = 125 × 15 = 1875 families
we expect 1875 families out of 2000 to have at least one boy.
how many have 2 boys ?
that is the same as the question in how many ways can I pick 2 elements out of 4.
that are 4 over 2 combinations
4! / (2! × (4-2)!) = 4!/(2!×2!) = 4×3/2 = 2×3 = 6
so, 6 out of the possible 16 possibilities have 2 boys.
the probabilty is therefore 6/16 = 3/8
and that applies as mean value of 2000 families to
2000 × 6/16 = 125 × 6 = 750 families
we expect 750 families out of 2000 to have two boys.
how many have 1 or 2 girls ?
there are 4 possibilities to have 1 girl (either the first, the second, the third or the fourth child).
there are (as before with the 2 boys) 6 possibilities to have 2 girls.
that is together 4+6=10 possibilities out of the 16 to have one or two girls.
the probabilty is therefore 10/16 = 5/8
and that applies as mean value of 2000 families to
2000 × 10/16 = 125 × 10 = 1250 families
we expect 1250 families out of 2000 to have one or two girls.
how many have no girls ?
that is the same as having only (4) boys.
there is only one possibility out of the 16 for that.
the probabilty is therefore 1/16.
and that applies as mean value of 2000 families to
2000 × 1/16 = 125 families
we expect 125 families out of 2000 to have no girls.