maths help!!!

the recovery ward in a maternity hospital has six beds. what is the probability that the mothers there have between them four girls and two boys?
(you may assume that there are no twins and that a baby is equally likely to be a girl or a boy)
how would you do this with binomial distribution?
im so confused

Duplicate thread.

Okay so you have 6 beds with a 50/50 chance that it's going to be a girl or boy and so there will be 2⁶ possibilities which equals 64.

So we know there are 64 possibilities.We know that if there are a 4 girls there is always going to be 2 boys and vice versa so for simplicity we will work out what is the probability of there being 2 boys so

P(Boys=2)= \frac{6C2}{64}

so that would be

\frac{6!}{2!*(6-2)!} = \frac{720}{48} = 15 \therefore \frac {15}{64}

of the time it will be 2 boys and 4 girls

Reply 3

help_me_learn

OP

15

Original post by GremlinIAMH

Okay so you have 6 beds with a 50/50 chance that it's going to be a girl or boy and so there will be 2⁶ possibilities which equals 64.

So we know there are 64 possibilities.We know that if there are a 4 girls there is always going to be 2 boys and vice versa so for simplicity we will work out what is the probability of there being 2 boys so

P(Boys=2)= \frac{6C2}{64}

so that would be

\frac{6!}{2!*(6-2)!} = \frac{720}{48} = 15 \therefore \frac {15}{64}

of the time it will be 2 boys and 4 girls

thank you so much!!

Reply 4

old_engineer

11

Original post by GremlinIAMH

Okay so you have 6 beds with a 50/50 chance that it's going to be a girl or boy and so there will be 2⁶ possibilities which equals 64.

So we know there are 64 possibilities.We know that if there are a 4 girls there is always going to be 2 boys and vice versa so for simplicity we will work out what is the probability of there being 2 boys so