Question

Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children. find the probability of exactly six boys in ten births. round the answer to the nearest thousandth.

Answer 1

You need to use the binomial probability formula:
$p(k) = \frac{n!}{k!(n-k)!} p^{k} (1-p)^{n-k}$

where:
n = total number of events = 10
k = number of events we are testing = 6
p = probability of event happening = 0.5

$p(6) = \frac{10!}{6!(10-6)!} 0.5^{6} (1-0.5)^{10-6}$
= 210×0.015625×0.0625
= 0.205078

Hence, the probability of getting 6 boys out of 10 births is p = 0.205.