I am trying to do my homework and I am stuck on this probability question. "If a family has eight kids what is the probability:
of all of them being the same gender?
having 4 boys and 4 girls?
and the genders to be split 5/3?

I am trying to do my homework and I am stuck on this probability question. "If a family has eight kids what is the probability:
of all of them being the same gender?

The probability of either gender is 1/2. Therefore, \frac{1}{2^{8}}



having 4 boys and 4 girls?


Use the binomial. C(8,4)(\frac{1}{2})^{4}(\frac{1}{2})^{4}



and the genders to be split 5/3?

You try this one. Okey-doke?

The probability of either gender is 1/2. Therefore, \frac{1}{2^{8}}



Not quite the answer to the question posed. This is the probability that the children are all boys. It's also the probability that they are all girls. It is not the probability that they are all the same gender. The prob that they are all the same gender is:

P(all boys or all girls) = P(all Boys) + P(all girls) - P (all boys and all girls).

Of course that last term is 0 for this problem (assuming we can ignore the possibility of hermaphrodites!) ;)

Not quite the answer to the question posed. This is the probability that the children are all boys. It's also the probability that they are all girls. It is not the probability that they are all the same gender. The prob that they are all the same gender is:

P(all boys or all girls) = P(all Boys) + P(all girls) - P (all boys and all girls).

Of course that last term is 0 for this problem (assuming we can ignore the possibility of hermaphrodites!) ;)


Yes, you are correct, ebaines. I had a brain fart. I was thinking the problem said all boys. :o