Assuming each birth can give life to a boy or girl with equal chance, then the probability of having a girl is
![(1)/(2)](https://img.qammunity.org/2019/formulas/mathematics/college/q5zg49mbtfrwobmmahi676fbgez56hhab0.png)
The probability of having exactly six girls is given by Bernoulli rule:
![\binom{8}{6}\left((1)/(2)\right)^6\left((1)/(2)\right)^2 = (8!)/(6!2!)\left((1)/(2)\right)^8 = (8\cdot 7)/(2) (1)/(2^8) = (28)/(2^8) \approx 0.11](https://img.qammunity.org/2019/formulas/mathematics/college/f60i5kapaym6jz2xqy721vxuc0u9ziufy9.png)
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((1)/(2)\right)^8 \left((1)/(2)\right)^0 + \binom{8}{1} \left((1)/(2)\right)^7 \left((1)/(2)\right)^1 + \binom{8}{2} \left((1)/(2)\right)^6 \left((1)/(2)\right)^2 = \left(\binom{8}{0} + \binom{8}{1} + \binom{8}{2}\right) (1)/(2^8)](https://img.qammunity.org/2019/formulas/mathematics/college/ptahcx8ylz5jy2vu4y7hbtwbr0e2y2hhsp.png)
which simplifies to
![((1+8+28))/(2^8) = (37)/(2^8) \approx 0.14](https://img.qammunity.org/2019/formulas/mathematics/college/ixb7oytsivzu5n7uurbz0pgkhsgdoi97qr.png)
Finally, I am not sure of what you mean with the point c, so please clarify the question.