Question

A. find the probability of getting exactly 6 girls in 8 births.

Answer 1

Assuming each birth can give life to a boy or girl with equal chance, then the probability of having a girl is $\frac{1}{2}$

The probability of having exactly six girls is given by Bernoulli rule:$\binom{8}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2 = \frac{8!}{6!2!}\left(\frac{1}{2}\right)^8 = \frac{8\cdot 7}{2} \frac{1}{2^8} = \frac{28}{2^8} \approx 0.11$

As for the second point, let's compute the probabilty of having zero, one or two boys, which is the same of having six or more girls, but requires less calculations:

$\binom{8}{0} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^0 + \binom{8}{1} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^1 + \binom{8}{2} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^2 = \left(\binom{8}{0} + \binom{8}{1} + \binom{8}{2}\right) \frac{1}{2^8}$

which simplifies to

$\frac{(1+8+28)}{2^8} = \frac{37}{2^8} \approx 0.14$

Finally, I am not sure of what you mean with the point c, so please clarify the question.