In a family of seven children, what is the probability of obtianing the following numbers of boys and girls?

Question

(a) all boys
(b) all children of the same sex
(c) six girls and one boy
(d) four boys and three girls
(e) four girls and three boys

If you guys get this, you're really smart. It took me forever to realize how easy this was. Goodluck.

Answer 1

a) 1/2. The probability of a boy or a girl does not depend on what came before. So, for each pregnancy, the probability is 1/2 that it will be a boy. For all seven to be boys the prob would be 1/2 ^7 = .0078 or .78%

b) Probability of all children the same sex. This is the same as above...all boys or all girls.

c) Six girls and one boy. You could have the following outcomes of six pregnancies, where (b) is boy and (g) is girl.

bbbbbbb (7 boys)
bbbbbbg (6 girls, one boy) ...etc.

You need to find all the combinations of 2 outcomes within 7 tries. You can use Pascal's triangle to find this, but it's basically:

(2)^7 which gives you 128 possible combinations. Of those combinations, the following are 6 girls and one boy:

ggggggb
gggggbg
ggggbgg
gggbggg
ggbgggg
gbggggg
bgggggg

As you can see, the chances are 7/128 or .054 or 5.4%

d / e) Once again, there are 128 possible arrangements of seven children.

You can work out the possible combinations--it will be the same probability for four girls and three boys, since the sex is irrelevant.

Regards,

Mysstere

Answer 2

(a) 7/7 now not that likelihood can certainly not be a million (b) all boys or all women=7/7+7/7 once more likelihood can certainly not be a complete quantity (c) 6/7*a million/7=6/forty nine (d) four/7*three/7=12/forty nine (e) four/7*three/7=12/forty nine