Question

. A family has n number of kids. The possibility of 2ⁿ combinations by gender has equal probability. In A, show that “the family has kids of both genders”. In B, please show “the family has at most one girl”. Please show that this event is dependent for n=2, but independent for n=3

Answer 1

A family has n number of kids.The possibility of 2^n combinations by gender has equal probability.

This 2^n is the all possible cases, because each of the n childs can either be a boy or a girl; so there are 2 options for each of the n kids.

So, all possible cases is 2^n.

A=Event that the family has kids of both genders.

Now, this event is complementary to the event that the family has n boys, or the family has n girls.

Both of these can occur in 1 way each.

So, we see

P(A)

=P(The family has kids of both genders)

=1-P(Family has only n boys 'or; Family has only n girls)

=1-P(Family has n boys)-P(Family has n girls)

This is equal to

So,

Now, B=Event that the family has at most one girl

Now,

P(B)

=P(Family has at most 1 girl)

=P(Family has 0 girls)+P(Family has 1 girl)

=P(Family has n boys)+P(Family has 1 girl)

Now, family can have n boys in only 1 way; and the family can have only 1 girl in n ways.

So,

Now,

A and B both occurs at the same time means the family has kids of both genders, and the family has at most 1 girl.

Now, this actually reduces to the family has 1 girl.

So, A and B means the family has exactly 1 girl.

So, we find that

Now, when n=2,

Now, as

So, A and B are not independent for n=2.

Now, for n=3

Now, in this case, we note that

So, for n=3, A and B are independent.