a couple plans to have 5 children. find the probability of having at least one girl?

Question

anyone know the steps to solving this? or the general rule?

Answer 1

Given 5 children there are 2^5 or 32 possible combinations of girls and boys. Since there is only one combination with all boys that leaves 31 combinations containing at least 1 girl. Therefore the probability is 31/32.

Answer 2

The couple may plan to have 5 children, but they may be able to have 0, 1, or 2 but not 3, 4, or 5.. Therefore the probability cannot be computed unless we are given data that let's us also compute the probability of the couple having 5 children.

Answer 3

This is a binomial distribution with the probability on any given birth of 1/2 it being a girl, with n=5 trials

Let x be then number of girls born

P(x>=1) = 1 - P(x=0)

P(x=o) = (5 chose 0)*(1/2)^0*(1/2)^5 = 1/32

P(x>=1) = 31/32

Answer 4

Assuming an equal chance of a girl or a boy, then the probability of having no girls in 5 children is:

(1/2)^5 = 1/32 = 0.03125, or just over 3%.

Therefore the probability of having at least one girl = 1-0.03125 =
0.96875. Almost a 97% probability.

Answer 5

Give the points to any of those first three whose answer you like. One further comment:

Any prob qu involving "at least one" is most easily solved by finding the probability of none -- as in this case solvers found the prob of all boys, which means no girls -- and then subtract from 1.