Respuesta :
Assuming each birth can give life to a boy or girl with equal chance, then the probability of having a girl is [tex] \frac{1}{2} [/tex]
The probability of having exactly six girls is given by Bernoulli rule:[tex] \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 [/tex]
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:
[tex] \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} [/tex]
which simplifies to
[tex] \frac{(1+8+28)}{2^8} = \frac{37}{2^8} \approx 0.14 [/tex]
Finally, I am not sure of what you mean with the point c, so please clarify the question.
