How many children should a family plan to have so that the probability of having at least one child of each se?

John C: How many children should a family plan to have so that the probability of having at least one child of each se?
How many children should a family plan to have so that the probability of
having at least one child of each sex is at least 0.95?

Answers and Views:

Answer by rauchers
1 – 0.5 ^(n-1) >=0.95

0.5^(n-1) >=0.05

ln 0.5^(n-1) >=ln 0.05

n-1 >= ln 0.05 / ln 0.5
n-1 >= 4.32
n> = 5.32
n=6

explaining
no mather what sex the first child has
the probability that secound child has the same sex is 0.5
and the probability that 2nd and 3th child have the same sex is 0.5^2
so the probability that all n childs have the same sex is
0.5^(n-1)
to have at least one from other sex
1- 0.5^(n-1)
and it must be greater or equal to 0.95
1- 0.5^(n-1) >=0.95

Answer by Northstar
For n children the probability of all of them being the same sex is:

(1/2)^(n – 1)
since it could be either all boys or all girls

Solve for the smallest integer n.

1 – 0.95 = 0.05 = 1/20

(1/2)^(n – 1) 1/20

For n = 6
(1/2)^(6 – 1) = (1/2)^5 = 1/32 < 1/20

The answer is 6 children.

Answer by MJ R
Because the odds are 50-50 for each, then odds are found by dividing the number of possibilities for all girls divided by the possibilities of all the other combined.

If you have 4 children, there is only one way to have four girls, there are
2^4 for all possible combinations.

possibility of all girls with 4= 1/(2^4) *100%=6.25%

Possibility of 5 all girls;

1/(2^5)*100%= 3.125%

Therefore you must have 5 children to have a better than 95% chance of having at least one of each sex.